Exploring Algebraic Identities

Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.

Answer

Let us take 4 consecutive square numbers. Let the consecutive integers be n, (n + 1), (n + 2) and (n + 3). Their squares are n², (n + 1)², (n + 2)² and (n + 3)².

Consider the operation: (sum of the smallest and largest squares) – (sum of the two middle squares).

⇒ [n² + (n + 3)²] - [(n + 1)² + (n + 2)²]

= [n² + n² + 6n + 9] - [n² + 2n + 1 + n² + 4n + 4]

= [2n² + 6n + 9] - [2n² + 6n + 5]

= 4.

Let us verify with an example. Take the consecutive squares 1, 4, 9, 16.

⇒ (1 + 16) - (4 + 9) = 17 - 13 = 4.

Now take 4, 9, 16, 25.

⇒ (4 + 25) - (9 + 16) = 29 - 25 = 4.

Hence, for any 4 consecutive square numbers, the difference between the sum of the smallest and the largest squares and the sum of the two middle squares is always 4.

What can you say about a and b if (a + b)² < a² + b² ?

Answer

We know the identity:

⇒ (a + b)² = a² + 2ab + b².

So,

⇒ (a + b)² - (a² + b²) = 2ab.

If (a + b)² < a² + b², then:

⇒ 2ab < 0

⇒ ab < 0.

This means a and b have opposite signs (one is positive and the other is negative).

Hence, (a + b)² < a² + b² when a and b have opposite signs.

What can you say about a and b if (a + b)² > a² + b² ?

Answer

We know:

⇒ (a + b)² - (a² + b²) = 2ab.

If (a + b)² > a² + b², then:

⇒ 2ab > 0

⇒ ab > 0.

This means a and b have the same sign (both positive or both negative).

Hence, (a + b)² > a² + b² when a and b have the same sign.

When will (a + b)² be equal to a² + b² ?

Did you observe that (a + b)² and a² + b² are both positive? What term will decide which is larger? Use the expansion of (a + b)² to decide.

Answer

We know:

⇒ (a + b)² - (a² + b²) = 2ab.

If (a + b)² = a² + b², then:

⇒ 2ab = 0

⇒ ab = 0.

This means at least one of a or b is equal to zero.

Hence, (a + b)² = a² + b² when at least one of a or b is zero.

From the expansion of (a + b)²:

⇒ (a + b)² = a² + 2ab + b².

Therefore:

⇒ (a + b)² - (a² + b²) = 2ab.

The sign of the term 2ab decides which is larger:

• If 2ab > 0, then (a + b)² > a² + b².
• If 2ab < 0, then (a + b)² < a² + b².
• If 2ab = 0, then (a + b)² = a² + b².

Hence, the term 2ab decides which of (a + b)² and a² + b² is larger.

Using the identity (a + b)² = a² + 2ab + b², expand the following:

(7x + 4y)²

Answer

Using the identity (a + b)² = a² + 2ab + b², with a = 7x and b = 4y:

⇒ (7x + 4y)² = (7x)² + 2(7x)(4y) + (4y)²

= 49x² + 56xy + 16y².

Hence, (7x + 4y)² = 49x² + 56xy + 16y².

Using the identity (a + b)² = a² + 2ab + b², expand the following:

$\left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2$

Answer

Using the identity (a + b)² = a² + 2ab + b², with $a = \dfrac{7}{5}x$ and $b = \dfrac{3}{2}y$:

$\Rightarrow \left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2 \\[1em] = \left(\dfrac{7}{5}x\right)^2 + 2\left(\dfrac{7}{5}x\right)\left(\dfrac{3}{2}y\right) + \left(\dfrac{3}{2}y\right)^2 \\[1em] = \dfrac{49}{25}x^2 + \dfrac{21}{5}xy + \dfrac{9}{4}y^2.$

Hence, $\left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2 = \dfrac{49}{25}x^2 + \dfrac{21}{5}xy + \dfrac{9}{4}y^2$.

Using the identity (a + b)² = a² + 2ab + b², expand the following:

(2.5p + 1.5q)²

Answer

Using the identity (a + b)² = a² + 2ab + b², with a = 2.5p and b = 1.5q:

⇒ (2.5p + 1.5q)² = (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²

= 6.25p² + 7.5pq + 2.25q².

Hence, (2.5p + 1.5q)² = 6.25p² + 7.5pq + 2.25q².

Using the identity (a + b)² = a² + 2ab + b², expand the following:

$\left(\dfrac{3}{4}s + 8t\right)^2$

Answer

Using the identity (a + b)² = a² + 2ab + b², with $a = \dfrac{3}{4}s$ and b = 8t:

$\Rightarrow \left(\dfrac{3}{4}s + 8t\right)^2 \\[1em] = \left(\dfrac{3}{4}s\right)^2 + 2\left(\dfrac{3}{4}s\right)(8t) + (8t)^2 \\[1em] = \dfrac{9}{16}s^2 + 12st + 64t^2.$

Hence, $\left(\dfrac{3}{4}s + 8t\right)^2 = \dfrac{9}{16}s^2 + 12st + 64t^2$.

Using the identity (a + b)² = a² + 2ab + b², expand the following:

$\left(x + \dfrac{1}{2y}\right)^2$

Answer

Using the identity (a + b)² = a² + 2ab + b², with a = x and $b = \dfrac{1}{2y}$:

$\Rightarrow \left(x + \dfrac{1}{2y}\right)^2 \\[1em] = x^2 + 2(x)\left(\dfrac{1}{2y}\right) + \left(\dfrac{1}{2y}\right)^2 \\[1em] = x^2 + \dfrac{x}{y} + \dfrac{1}{4y^2}.$

Hence, $\left(x + \dfrac{1}{2y}\right)^2 = x^2 + \dfrac{x}{y} + \dfrac{1}{4y^2}$.

Using the identity (a + b)² = a² + 2ab + b², expand the following:

$\left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2$

Answer

Using the identity (a + b)² = a² + 2ab + b², with $a = \dfrac{1}{x}$ and $b = \dfrac{1}{y}$:

$\Rightarrow \left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2 \\[1em] = \left(\dfrac{1}{x}\right)^2 + 2\left(\dfrac{1}{x}\right)\left(\dfrac{1}{y}\right) + \left(\dfrac{1}{y}\right)^2 \\[1em] = \dfrac{1}{x^2} + \dfrac{2}{xy} + \dfrac{1}{y^2}.$

Hence, $\left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2 = \dfrac{1}{x^2} + \dfrac{2}{xy} + \dfrac{1}{y^2}$.

Using the same identity, find the values of the following:

(64)²

Answer

Writing 64 as (60 + 4) and using (a + b)² = a² + 2ab + b²:

⇒ (64)² = (60 + 4)² = (60)² + 2(60)(4) + (4)²

= 3600 + 480 + 16

= 4096.

Hence, (64)² = 4096.

Using the same identity, find the values of the following:

(105)²

Answer

Writing 105 as (100 + 5) and using (a + b)² = a² + 2ab + b²:

⇒ (105)²

= (100 + 5)²

= (100)² + 2(100)(5) + (5)²

= 10000 + 1000 + 25

= 11025.

Hence, (105)² = 11025.

Using the same identity, find the values of the following:

(205)²

Answer

Writing 205 as (200 + 5) and using (a + b)² = a² + 2ab + b²:

⇒ (205)² = (200 + 5)² = (200)² + 2(200)(5) + (5)²

= 40000 + 2000 + 25

= 42025.

Hence, (205)² = 42025.

What if we replace b by –b in (a + b)² = a² + 2ab + b² ?

Answer

Replacing b by –b in the identity (a + b)² = a² + 2ab + b²:

⇒ [a + (-b)]² = a² + 2a(-b) + (-b)²

⇒ (a - b)² = a² - 2ab + b².

Hence, replacing b by –b in (a + b)² = a² + 2ab + b², we obtain the new identity (a – b)² = a² – 2ab + b².

Factor completely:

9x² + 24xy + 16y²

Answer

Rewriting the expression in the form a² + 2ab + b²:

⇒ 9x² + 24xy + 16y²

= (3x)² + 2(3x)(4y) + (4y)²

= (3x + 4y)².

Hence, 9x² + 24xy + 16y² = (3x + 4y)².

Factor completely:

4s² + 20st + 25t²

Answer

Rewriting the expression in the form a² + 2ab + b²:

⇒ 4s² + 20st + 25t²

= (2s)² + 2(2s)(5t) + (5t)²

= (2s + 5t)².

Hence, 4s² + 20st + 25t² = (2s + 5t)².

Factor completely:

49x² + 28xy + 4y²

Answer

Rewriting the expression in the form a² + 2ab + b²:

⇒ 49x² + 28xy + 4y²

= (7x)² + 2(7x)(2y) + (2y)²

= (7x + 2y)².

Hence, 49x² + 28xy + 4y² = (7x + 2y)².

Factor completely:

$64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2$

Answer

Rewriting the expression in the form a² + 2ab + b²:

$\Rightarrow 64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2 \\[1em] = (8p)^2 + 2(8p)\left(\dfrac{2}{3}q\right) + \left(\dfrac{2}{3}q\right)^2 \\[1em] = \left(8p + \dfrac{2}{3}q\right)^2.$

Hence, $64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2 = \left(8p + \dfrac{2}{3}q\right)^2$.

Factor completely:

$3a^2 + 4ab + \dfrac{4}{3}b^2$

Answer

Taking $\dfrac{1}{3}$ as a common factor:

$\Rightarrow 3a^2 + 4ab + \dfrac{4}{3}b^2 \\[1em] = \dfrac{1}{3}\left(9a^2 + 12ab + 4b^2\right) \\[1em] = \dfrac{1}{3}\left[(3a)^2 + 2(3a)(2b) + (2b)^2\right] \\[1em] = \dfrac{1}{3}(3a + 2b)^2.$

Hence, $3a^2 + 4ab + \dfrac{4}{3}b^2 = \dfrac{1}{3}(3a + 2b)^2$.

Factor completely:

$\dfrac{9}{5}s^2 + 6sv + 5v^2$

Answer

Taking $\dfrac{1}{5}$ as a common factor:

$\Rightarrow \dfrac{9}{5}s^2 + 6sv + 5v^2 \\[1em] = \dfrac{1}{5}\left(9s^2 + 30sv + 25v^2\right) \\[1em] = \dfrac{1}{5}\left[(3s)^2 + 2(3s)(5v) + (5v)^2\right] \\[1em] = \dfrac{1}{5}(3s + 5v)^2.$

Hence, $\dfrac{9}{5}s^2 + 6sv + 5v^2 = \dfrac{1}{5}(3s + 5v)^2$.

Find the values of the following using the identity (a – b)² = a² – 2ab + b².

(79)²

Answer

Writing 79 as (80 – 1) and using (a – b)² = a² – 2ab + b²:

⇒ (79)²

= (80 - 1)² = (80)² - 2(80)(1) + (1)²

= 6400 - 160 + 1

= 6241.

Hence, (79)² = 6241.

Find the values of the following using the identity (a – b)² = a² – 2ab + b².

(193)²

Answer

Writing 193 as (200 – 7) and using (a – b)² = a² – 2ab + b²:

⇒ (193)²

= (200 - 7)²

= (200)² - 2(200)(7) + (7)²

= 40000 - 2800 + 49

= 37249.

Hence, (193)² = 37249.

Find the values of the following using the identity (a – b)² = a² – 2ab + b².

(299)²

Answer

Writing 299 as (300 – 1) and using (a – b)² = a² – 2ab + b²:

⇒ (299)²

= (300 - 1)²

= (300)² - 2(300)(1) + (1)²

= 90000 - 600 + 1

= 89401.

Hence, (299)² = 89401.

Label the squares and rectangles in Fig. 4.4 so that it represents the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Answer

The figure shows a square of side (a + b + c) units divided into 9 regions — 3 squares and 6 rectangles. Labelling each region with its area:

Adding the areas of all 9 regions:

⇒ (a + b + c)² = a² + ab + ac + ab + b² + bc + ac + bc + c²

= a² + b² + c² + 2ab + 2bc + 2ca.

Hence, the geometrical model confirms the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

117²

Answer

Writing 117 as (120 – 3) and using (a – b)² = a² – 2ab + b²:

⇒ (117)²

= (120 - 3)²

= (120)² - 2(120)(3) + (3)²

= 14400 - 720 + 9

= 13689.

Hence, (117)² = 13689.

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

78²

Answer

Writing 78 as (80 – 2) and using (a – b)² = a² – 2ab + b²:

⇒ (78)²

= (80 - 2)²

= (80)² - 2(80)(2) + (2)²

= 6400 - 320 + 4

= 6084.

Hence, (78)² = 6084.

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

198²

Answer

Writing 198 as (200 – 2) and using (a – b)² = a² – 2ab + b²:

⇒ (198)²

= (200 - 2)²

= (200)² - 2(200)(2) + (2)²

= 40000 - 800 + 4

= 39204.

Hence, (198)² = 39204.

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

214²

Answer

Writing 214 as (200 + 14) and using (a + b)² = a² + 2ab + b²:

⇒ (214)²

= (200 + 14)²

= (200)² + 2(200)(14) + (14)²

= 40000 + 5600 + 196

= 45796.

Hence, (214)² = 45796.

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

1104²

Answer

Writing 1104 as (1100 + 4) and using (a + b)² = a² + 2ab + b²:

⇒ (1104)²

= (1100 + 4)²

= (1100)² + 2(1100)(4) + (4)²

= 1210000 + 8800 + 16

= 1218816.

Hence, (1104)² = 1218816.

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

1120²

Answer

Writing 1120 as (1000 + 100 + 20) and using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:

⇒ (1120)²

= (1000 + 100 + 20)²

= (1000)² + (100)² + (20)² + 2(1000)(100) + 2(100)(20) + 2(1000)(20)

= 1000000 + 10000 + 400 + 200000 + 4000 + 40000

= 1254400.

Hence, (1120)² = 1254400.

Factor using suitable identities:

16y² – 24y + 9

Answer

Rewriting the expression in the form a² – 2ab + b²:

⇒ 16y² - 24y + 9

= (4y)² - 2(4y)(3) + (3)²

= (4y - 3)².

Hence, 16y² – 24y + 9 = (4y – 3)².

Factor using suitable identities:

$\dfrac{9}{4}s^2 + 6st + 4t^2$

Answer

Rewriting the expression in the form a² + 2ab + b²:

$\Rightarrow \dfrac{9}{4}s^2 + 6st + 4t^2 \\[1em] = \left(\dfrac{3}{2}s\right)^2 + 2\left(\dfrac{3}{2}s\right)(2t) + (2t)^2 \\[1em] = \left(\dfrac{3}{2}s + 2t\right)^2.$

Hence, $\dfrac{9}{4}s^2 + 6st + 4t^2 = \left(\dfrac{3}{2}s + 2t\right)^2$.

Factor using suitable identities:

$\dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2$

Answer

Rewriting the expression in the form a² + b² + c² + 2ab + 2bc + 2ca:

$\Rightarrow \dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2 \\[1em] = \left(\dfrac{m}{3}\right)^2 + \left(\dfrac{k}{2}\right)^2 + (3n)^2 + 2\left(\dfrac{m}{3}\right)\left(\dfrac{k}{2}\right) + 2\left(\dfrac{k}{2}\right)(3n) + 2\left(\dfrac{m}{3}\right)(3n) \\[1em] = \left(\dfrac{m}{3} + \dfrac{k}{2} + 3n\right)^2.$

Hence, $\dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2 = \left(\dfrac{m}{3} + \dfrac{k}{2} + 3n\right)^2$.

Factor using suitable identities:

$\dfrac{p^2}{16} - 2 + \dfrac{16}{p^2}$

Answer

Rewriting the expression in the form a² – 2ab + b²:

$\Rightarrow \dfrac{p^2}{16} - 2 + \dfrac{16}{p^2} \\[1em] = \left(\dfrac{p}{4}\right)^2 - 2\left(\dfrac{p}{4}\right)\left(\dfrac{4}{p}\right) + \left(\dfrac{4}{p}\right)^2 \\[1em] = \left(\dfrac{p}{4} - \dfrac{4}{p}\right)^2.$

Hence, $\dfrac{p^2}{16} - 2 + \dfrac{16}{p^2} = \left(\dfrac{p}{4} - \dfrac{4}{p}\right)^2$.

Factor using suitable identities:

9a² + 4b² + c² – 12ab + 6ac – 4bc

Answer

Rewriting the expression in the form A² + B² + C² + 2AB + 2BC + 2CA, with A = 3a, B = –2b, C = c:

⇒ 9a² + 4b² + c² - 12ab + 6ac - 4bc

= (3a)² + (-2b)² + (c)² + 2(3a)(-2b) + 2(-2b)(c) + 2(3a)(c)

= (3a - 2b + c)².

Hence, 9a² + 4b² + c² – 12ab + 6ac – 4bc = (3a – 2b + c)².

Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:

(p + 3q + 7r)²

Answer

Using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca, with a = p, b = 3q, c = 7r:

⇒ (p + 3q + 7r)² = p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(p)(7r)

= p² + 9q² + 49r² + 6pq + 42qr + 14pr.

Hence, (p + 3q + 7r)² = p² + 9q² + 49r² + 6pq + 42qr + 14pr.

Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:

(3x – 2y + 4z)²

Answer

Using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca, with a = 3x, b = –2y, c = 4z:

⇒ (3x - 2y + 4z)² = (3x)² + (-2y)² + (4z)² + 2(3x)(-2y) + 2(-2y)(4z) + 2(3x)(4z)

= 9x² + 4y² + 16z² - 12xy - 16yz + 24xz.

Hence, (3x – 2y + 4z)² = 9x² + 4y² + 16z² – 12xy – 16yz + 24xz.

Is this an identity?

(a + b – c)² + (a – b + c)² + (a – b – c)² = 2a² + 2b² + 2c².

Answer

Expanding the L.H.S. using the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx:

⇒ (a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ac.

⇒ (a - b + c)² = a² + b² + c² - 2ab - 2bc + 2ac.

⇒ (a - b - c)² = a² + b² + c² - 2ab + 2bc - 2ac.

Adding these three expressions:

⇒ (a + b - c)² + (a - b + c)² + (a - b - c)²

= 3a² + 3b² + 3c² + (2ab - 2ab - 2ab) + (-2bc - 2bc + 2bc) + (-2ac + 2ac - 2ac)

= 3a² + 3b² + 3c² - 2ab - 2bc - 2ac.

Since L.H.S. = 3a² + 3b² + 3c² – 2ab – 2bc – 2ac ≠ 2a² + 2b² + 2c² = R.H.S., the given equation is not true for all values of a, b, c.

To verify, let a = 1, b = 1, c = 1:

⇒ L.H.S. = (1 + 1 - 1)² + (1 - 1 + 1)² + (1 - 1 - 1)² = 1 + 1 + 1 = 3.

⇒ R.H.S. = 2(1)² + 2(1)² + 2(1)² = 6.

Since L.H.S. ≠ R.H.S., the given equation is not an identity.

Hence, (a + b – c)² + (a – b + c)² + (a – b – c)² = 2a² + 2b² + 2c² is not an identity.

Try to evaluate the following using a suitable identity:

(i) 35²

(ii) 65²

(iii) 85²

(iv) 105²

Do you observe any interesting pattern?

Answer

Using the identity a² = (a + b)(a – b) + b², with b = 5:

(i) 35²:

⇒ 35² = (35 + 5)(35 - 5) + 5²

= 40 × 30 + 25

= 1200 + 25

= 1225.

(ii) 65²:

⇒ 65² = (65 + 5)(65 - 5) + 5²

= 70 × 60 + 25

= 4200 + 25

= 4225.

(iii) 85²:

⇒ 85² = (85 + 5)(85 - 5) + 5²

= 90 × 80 + 25

= 7200 + 25

= 7225.

(iv) 105²:

⇒ 105² = (105 + 5)(105 - 5) + 5²

= 110 × 100 + 25

= 11000 + 25

= 11025.

Pattern observed: For any number ending in 5, say (10n + 5), the square is:

⇒ (10n + 5)² = n(n + 1) × 100 + 25.

So the square always ends in 25 and the digits before 25 are the product n(n + 1), where n is the digit(s) before the 5.

• 35²: n = 3, so n(n + 1) = 3 × 4 = 12, giving 1225.
• 65²: n = 6, so n(n + 1) = 6 × 7 = 42, giving 4225.
• 85²: n = 8, so n(n + 1) = 8 × 9 = 72, giving 7225.
• 105²: n = 10, so n(n + 1) = 10 × 11 = 110, giving 11025.

Hence, the square of a number ending in 5 always ends in 25, and the digits before 25 are n(n + 1), where n is the digit(s) before the 5.

Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.

Answer

The top row of the figure consists of four squares with sides (a + b + c), (a + b – c), (a – b + c) and (a – b – c) respectively. The bottom row consists of three squares with sides 2a, 2b and 2c respectively.

The sum of the areas of the squares in the top row equals the sum of the areas of the squares in the bottom row:

⇒ (a + b + c)² + (a + b - c)² + (a - b + c)² + (a - b - c)² = (2a)² + (2b)² + (2c)².

Let us verify by expanding the left side:

⇒ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac.

⇒ (a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ac.

⇒ (a - b + c)² = a² + b² + c² - 2ab - 2bc + 2ac.

⇒ (a - b - c)² = a² + b² + c² - 2ab + 2bc - 2ac.

Adding them:

⇒ Sum = 4a² + 4b² + 4c²

= (2a)² + (2b)² + (2c)².

Hence, the identity represented is (a + b + c)² + (a + b – c)² + (a – b + c)² + (a – b – c)² = 4(a² + b² + c²).

Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.

Answer

For the expression x² + 7x + 12, we need two numbers a and b such that a + b = 7 (the coefficient of x) and ab = 12 (the constant term).

Let us check the splits of 7x:

• 2x + 5x: Here a = 2 and b = 5. Then a + b = 7 but ab = 10 ≠ 12. So a rectangular arrangement cannot be formed.

• 1x + 6x: Here a = 1 and b = 6. Then a + b = 7 but ab = 6 ≠ 12. So this also does not work.

• 3x + 4x: Here a = 3 and b = 4. Then a + b = 7 and ab = 12. So this is the only split that forms a valid rectangular arrangement.

Hence, only the split 7x = 3x + 4x allows a rectangular arrangement to be formed for x² + 7x + 12, because both a + b = 7 and ab = 12 must be satisfied simultaneously.

Figure out the product of x + 2 and x + 3 using algebra tiles.

Answer

To find the product (x + 2)(x + 3) using algebra tiles, we represent x + 2 as an x-tile and 2 unit tiles, and x + 3 as an x-tile and 3 unit tiles.

The product forms a rectangle with length (x + 3) and breadth (x + 2). This rectangle contains:

• 1 x²-tile (representing x²)
• 3 x-tiles along the length (representing 3x)
• 2 x-tiles along the breadth (representing 2x)
• 6 unit tiles arranged in a 2 × 3 array (representing 6)

Total area:

⇒ (x + 2)(x + 3) = x² + 3x + 2x + 6

= x² + 5x + 6.

Hence, (x + 2)(x + 3) = x² + 5x + 6.

Lay out algebra tiles for x² + 11x + 30 in such a way that you will see its factors.

Answer

To factor x² + 11x + 30, we need to find two numbers a and b such that a + b = 11 and ab = 30.

Considering factor pairs of 30: (1, 30), (2, 15), (3, 10), (5, 6).

The pair (5, 6) satisfies both conditions: 5 + 6 = 11 and 5 × 6 = 30 .

So we split 11x as 5x + 6x. The algebra tiles arrangement consists of:

• 1 x²-tile
• 5 x-tiles arranged along one side of the x²-tile
• 6 x-tiles arranged along the other side of the x²-tile
• 30 unit tiles arranged in a 5 × 6 array

This forms a rectangle of dimensions (x + 5) and (x + 6):

⇒ x² + 11x + 30 = x² + 5x + 6x + 30

= x(x + 6) + 5(x + 6)

= (x + 5)(x + 6).

Hence, x² + 11x + 30 = (x + 5)(x + 6).

We have seen that (x + 3)(x + 4) = x² + 7x + 12.

Also (x + 6)(x + 7) = x² + 13x + 42.

Generalise the pattern to get an expression for (x + a)(x + b).

Answer

Observing the two given identities:

• In (x + 3)(x + 4) = x² + 7x + 12: here 7 = 3 + 4 and 12 = 3 × 4.
• In (x + 6)(x + 7) = x² + 13x + 42: here 13 = 6 + 7 and 42 = 6 × 7.

Generalising for (x + a)(x + b) using the distributive property:

⇒ (x + a)(x + b) = x(x + b) + a(x + b)

= x² + bx + ax + ab

= x² + (a + b)x + ab.

Hence, (x + a)(x + b) = x² + (a + b)x + ab.

Fill in the blanks to complete the following identity:

s² – 11s + 24 = ( ............... ) ( ............... )

Answer

For s² – 11s + 24, we need numbers a and b such that a + b = –11 and ab = 24.

Choosing a = –3 and b = –8: a + b = –11 and ab = 24 .

⇒ s² - 11s + 24 = s² - 3s - 8s + 24

= s(s - 3) - 8(s - 3)

= (s - 3)(s - 8).

Hence, s² – 11s + 24 = (s – 3)(s – 8).

Fill in the blanks to complete the following identity:

( ............... ) (x + 1) = (3x² – 4x – 7)

Answer

Dividing 3x² – 4x – 7 by (x + 1):

We need to factor 3x² – 4x – 7. We need two numbers whose product is 3 × (–7) = –21 and whose sum is –4. The numbers are –7 and 3.

⇒ 3x² - 4x - 7 = 3x² - 7x + 3x - 7

= x(3x - 7) + 1(3x - 7)

= (3x - 7)(x + 1).

Hence, (3x – 7)(x + 1) = 3x² – 4x – 7.

Fill in the blanks to complete the following identity:

10x² – 11x – 6 = (2x – ............... ) ( ............... + 2)

Answer

For 10x² – 11x – 6, we need two numbers whose product is 10 × (–6) = –60 and whose sum is –11. The numbers are –15 and 4.

⇒ 10x² - 11x - 6 = 10x² - 15x + 4x - 6

= 5x(2x - 3) + 2(2x - 3)

= (2x - 3)(5x + 2).

Hence, 10x² – 11x – 6 = (2x – 3)(5x + 2).

Fill in the blanks to complete the following identity:

6x² + 7x + 2 = ( ............... ) ( ............... )

Answer

For 6x² + 7x + 2, we need two numbers whose product is 6 × 2 = 12 and whose sum is 7. The numbers are 3 and 4.

⇒ 6x² + 7x + 2 = 6x² + 3x + 4x + 2

= 3x(2x + 1) + 2(2x + 1)

= (2x + 1)(3x + 2).

Hence, 6x² + 7x + 2 = (2x + 1)(3x + 2).

Select and use the identity that will help you to find the following products without multiplying directly:

(41)²

Answer

Writing 41 as (40 + 1) and using (a + b)² = a² + 2ab + b²:

⇒ (41)² = (40 + 1)² = (40)² + 2(40)(1) + (1)²

= 1600 + 80 + 1

= 1681.

Hence, (41)² = 1681.

Select and use the identity that will help you to find the following products without multiplying directly:

(27)²

Answer

Writing 27 as (30 – 3) and using (a – b)² = a² – 2ab + b²:

⇒ (27)² = (30 - 3)² = (30)² - 2(30)(3) + (3)²

= 900 - 180 + 9

= 729.

Hence, (27)² = 729.

Select and use the identity that will help you to find the following products without multiplying directly:

(23 × 17)

Answer

Writing 23 as (20 + 3) and 17 as (20 – 3), and using (a + b)(a – b) = a² – b²:

⇒ 23 × 17 = (20 + 3)(20 - 3) = (20)² - (3)²

= 400 - 9

= 391.

Hence, 23 × 17 = 391.

Select and use the identity that will help you to find the following products without multiplying directly:

(135)²

Answer

Writing 135 as (100 + 30 + 5) and using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:

⇒ (135)² = (100 + 30 + 5)²

= (100)² + (30)² + (5)² + 2(100)(30) + 2(30)(5) + 2(100)(5)

= 10000 + 900 + 25 + 6000 + 300 + 1000

= 18225.

Hence, (135)² = 18225.

Select and use the identity that will help you to find the following products without multiplying directly:

(97)²

Answer

Writing 97 as (100 – 3) and using (a – b)² = a² – 2ab + b²:

⇒ (97)² = (100 - 3)² = (100)² - 2(100)(3) + (3)²

= 10000 - 600 + 9

= 9409.

Hence, (97)² = 9409.

Select and use the identity that will help you to find the following products without multiplying directly:

(18 × 29)

Answer

Writing 18 as (23 – 5) and 29 as (23 + 6), and using (x + a)(x + b) = x² + (a + b)x + ab, with x = 23, a = –5, b = 6:

⇒ 18 × 29 = (23 - 5)(23 + 6)

= (23)² + (-5 + 6)(23) + (-5)(6)

= 529 + 23 - 30

= 522.

Hence, 18 × 29 = 522.

Select and use the identity that will help you to find the following products without multiplying directly:

(34 × 43)

Answer

Writing 34 as (40 – 6) and 43 as (40 + 3), and using (x + a)(x + b) = x² + (a + b)x + ab, with x = 40, a = –6, b = 3:

⇒ 34 × 43 = (40 - 6)(40 + 3)

= (40)² + (-6 + 3)(40) + (-6)(3)

= 1600 - 120 - 18

= 1462.

Hence, 34 × 43 = 1462.

Select and use the identity that will help you to find the following products without multiplying directly:

(205)²

Answer

Writing 205 as (200 + 5) and using (a + b)² = a² + 2ab + b²:

⇒ (205)² = (200 + 5)² = (200)² + 2(200)(5) + (5)²

= 40000 + 2000 + 25

= 42025.

Hence, (205)² = 42025.

Factor the following:

9a² + b² + 4c² – 6ab + 12ac – 4bc

Answer

Rewriting the expression in the form A² + B² + C² + 2AB + 2BC + 2CA, with A = 3a, B = –b, C = 2c:

⇒ 9a² + b² + 4c² - 6ab + 12ac - 4bc

= (3a)² + (-b)² + (2c)² + 2(3a)(-b) + 2(-b)(2c) + 2(3a)(2c)

= (3a - b + 2c)².

Hence, 9a² + b² + 4c² – 6ab + 12ac – 4bc = (3a – b + 2c)².

Factor the following:

16s² + 25t² – 40st

Answer

Rewriting the expression in the form a² – 2ab + b²:

⇒ 16s² + 25t² - 40st = (4s)² - 2(4s)(5t) + (5t)²

= (4s - 5t)².

Hence, 16s² + 25t² – 40st = (4s – 5t)².

Factor the following:

r² – r – 42

Answer

For r² – r – 42, we need two numbers whose sum is –1 and whose product is –42. The numbers are –7 and 6.

⇒ r² - r - 42 = r² - 7r + 6r - 42

= r(r - 7) + 6(r - 7)

= (r - 7)(r + 6).

Hence, r² – r – 42 = (r – 7)(r + 6).

Factor the following:

49g² + 14gh + h²

Answer

Rewriting the expression in the form a² + 2ab + b²:

⇒ 49g² + 14gh + h² = (7g)² + 2(7g)(h) + h²

= (7g + h)².

Hence, 49g² + 14gh + h² = (7g + h)².

Factor the following:

64u² + 121v² + 4w² – 176uv – 32uw + 44vw

Answer

Rewriting the expression in the form A² + B² + C² + 2AB + 2BC + 2CA, with A = 8u, B = –11v, C = –2w:

⇒ 64u² + 121v² + 4w² - 176uv - 32uw + 44vw

= (8u)² + (-11v)² + (-2w)² + 2(8u)(-11v) + 2(-11v)(-2w) + 2(8u)(-2w)

= (8u - 11v - 2w)².

Hence, 64u² + 121v² + 4w² – 176uv – 32uw + 44vw = (8u – 11v – 2w)².

James and Reshma were talking about algebraic identities they learnt in school.

James: (a – b)² (a + b) = (a² – 2ab + b²)(a + b)

Reshma: I have a different idea. (a – b)² (a + b) = (a – b) [(a – b) (a + b)] = (a – b)(a² – b²)

I will find this product to get the answer.

According to you, who is correct and why?

Try to combine more such identities and find new results.

Answer

Let us evaluate both methods.

James's method:

⇒ (a - b)² (a + b) = (a² - 2ab + b²)(a + b)

= a³ + a²b - 2a²b - 2ab² + ab² + b³

= a³ - a²b - ab² + b³.

Reshma's method:

⇒ (a - b)² (a + b) = (a - b)[(a - b)(a + b)]

= (a - b)(a² - b²)

= a³ - ab² - a²b + b³

= a³ - a²b - ab² + b³.

Both methods give the same answer: a³ – a²b – ab² + b³.

Both are correct. They are simply two different approaches that lead to the same simplified expression. Reshma's approach is more elegant as it uses the identity (a – b)(a + b) = a² – b² to reduce the steps.

By combining identities, we can derive new ones. For example:

⇒ (a + b)² (a - b) = (a + b)[(a + b)(a - b)]

= (a + b)(a² - b²)

= a³ - ab² + a²b - b³

= a³ + a²b - ab² - b³.

Hence, both James and Reshma are correct as they arrive at the same expression a³ – a²b – ab² + b³, although by different methods.

We already know that x² – y² = (x – y)(x + y).

Further, we have verified that x³ – y³ = (x – y)(x² + xy + y²).

Observe that x – y is a common factor of x² – y² and x³ – y³.

Do you think x – y is also a factor of x⁴ – y⁴ ?

Note that x⁴ – y⁴ = (x²)² – (y²)² = (x² – y²) (x² + y²).

Can you see how x – y is a factor of x⁴ – y⁴ ?

How about x⁵ – y⁵? Does this also have x – y as a factor?

Answer

For x⁴ – y⁴:

⇒ x⁴ – y⁴ = (x²)² – (y²)² = (x² – y²)(x² + y²)

= (x – y)(x + y)(x² + y²).

So yes, (x – y) is a factor of x⁴ – y⁴.

For x⁵ – y⁵:

By direct expansion, we can show that:

⇒ x⁵ – y⁵ = (x – y)(x⁴ + x³y + x²y² + xy³ + y⁴).

Let us verify by multiplying:

⇒ (x – y)(x⁴ + x³y + x²y² + xy³ + y⁴)

= x⁵ + x⁴y + x³y² + x²y³ + xy⁴ – x⁴y – x³y² – x²y³ – xy⁴ – y⁵

= x⁵ – y⁵.

So yes, (x – y) is a factor of x⁵ – y⁵.

In general, xⁿ – yⁿ = (x – y)(x^n–1 + x^n–2y + … + xy^n–2 + y^n–1) for every natural number n.

Hence, (x – y) is a factor of xⁿ – yⁿ for every natural number n, including x⁴ – y⁴ and x⁵ – y⁵.

Try to simplify the following rational expression:

$\dfrac{36s^2 - 12st + t^2}{t^2 + 2ts - 48s^2} = \dfrac{(6s - t)^2}{(.......... + ..........)(.......... + ..........)}$

Answer

Factoring the numerator:

⇒ 36s² - 12st + t²

= (6s)² - 2(6s)(t) + t²

= (6s - t)².

Factoring the denominator:

For t² + 2ts – 48s², we need two terms (in terms of s) whose sum is 2s and whose product is –48s². The terms are 8s and –6s.

⇒ t² + 2ts - 48s² = t² + 8ts - 6ts - 48s²

= t(t + 8s) - 6s(t + 8s)

= (t + 8s)(t - 6s).

So:

$\Rightarrow \dfrac{36s^2 - 12st + t^2}{t^2 + 2ts - 48s^2} = \dfrac{(6s - t)^2}{(t + 8s)(t - 6s)}.$

Simplifying further, since (6s – t) = –(t – 6s):

$\Rightarrow \dfrac{(6s - t)^2}{(t + 8s)(t - 6s)} = \dfrac{(6s - t)(6s - t)}{(t + 8s)\big(-(6s - t)\big)} \\[1em] = \dfrac{-(6s - t)}{t + 8s} = \dfrac{t - 6s}{t + 8s}.$

Hence, $\dfrac{36s^2 - 12st + t^2}{t^2 + 2ts - 48s^2} = \dfrac{(6s - t)^2}{(t + 8s)(t - 6s)} = \dfrac{t - 6s}{t + 8s}$.

Simplify the following rational expression assuming that the expression in the denominator is not equal to zero:

$\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$

Answer

Factoring the numerator:

⇒ 3p² - 3pq - 18q² = 3(p² - pq - 6q²)

= 3(p² - 3pq + 2pq - 6q²)

= 3[p(p - 3q) + 2q(p - 3q)]

= 3(p - 3q)(p + 2q).

Factoring the denominator:

⇒ p² + 3pq - 10q² = p² + 5pq - 2pq - 10q²

= p(p + 5q) - 2q(p + 5q)

= (p + 5q)(p - 2q).

So:

$\Rightarrow \dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} = \dfrac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}.$

Hence, $\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} = \dfrac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}$.

Simplify the following rational expression assuming that the expression in the denominator is not equal to zero:

$\dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}$

Answer

Factoring the numerator using (a – b)³ = a³ – 3a²b + 3ab² – b³:

⇒ n³ - 3n²m + 3nm² - m³ = (n - m)³.

Factoring the denominator:

⇒ 5m² - 10mn + 5n² = 5(m² - 2mn + n²)

= 5(m - n)² = 5(n - m)².

So:

$\Rightarrow \dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2} \\[1em] = \dfrac{(n - m)^3}{5(n - m)^2} \\[1em] = \dfrac{n - m}{5}.$

Hence, $\dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2} = \dfrac{n - m}{5}$.

Simplify the following rational expression assuming that the expression in the denominator is not equal to zero:

$\dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}$

Answer

Factoring the numerator using the identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca), with a = w, b = –v, c = x:

⇒ w³ + (-v)³ + x³ - 3(w)(-v)(x)

= w³ - v³ + x³ + 3wvx.

So:

⇒ w³ - v³ + x³ + 3wvx

= (w - v + x)(w² + v² + x² + wv + vx - wx).

Factoring the denominator:

⇒ w² + v² + x² - 2wv - 2vx + 2wx

= w² + (-v)² + x² + 2(w)(-v) + 2(-v)(x) + 2(w)(x)

= (w - v + x)².

So:

$\Rightarrow \dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx} \\[1em] = \dfrac{(w - v + x)(w^2 + v^2 + x^2 + wv + vx - wx)}{(w - v + x)^2} \\[1em] = \dfrac{w^2 + v^2 + x^2 + wv + vx - wx}{w - v + x}.$

Hence, $\dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx} = \dfrac{w^2 + v^2 + x^2 + wv + vx - wx}{w - v + x}$.

Simplify the following rational expression assuming that the expression in the denominator is not equal to zero:

$\dfrac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)}$

Answer

Factoring the numerator:

⇒ 4y² - 20yz + 25z²

= (2y)² - 2(2y)(5z) + (5z)²

= (2y - 5z)².

Factoring the denominator:

⇒ 25z² - 4y² = (5z)² - (2y)²

= (5z - 2y)(5z + 2y).

Since (5z – 2y) = –(2y – 5z):

$\Rightarrow \dfrac{4y^2 - 20yz + 25z^2}{25z^2 - 4y^2} = \dfrac{(2y - 5z)^2}{(5z - 2y)(5z + 2y)} \\[1em] = \dfrac{(2y - 5z)(2y - 5z)}{-(2y - 5z)(5z + 2y)} \\[1em] = \dfrac{-(2y - 5z)}{5z + 2y} \\[1em] = \dfrac{5z - 2y}{5z + 2y}.$

Hence, $\dfrac{4y^2 - 20yz + 25z^2}{25z^2 - 4y^2} = \dfrac{5z - 2y}{5z + 2y}$.

Simplify the following rational expression assuming that the expression in the denominator is not equal to zero:

$\dfrac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}$

Answer

Factoring each quadratic:

⇒ x² + x - 6 = x² + 3x - 2x - 6 = (x + 3)(x - 2).

⇒ x² - 7x + 12 = x² - 3x - 4x + 12 = (x - 3)(x - 4).

⇒ x² - 6x + 8 = x² - 2x - 4x + 8 = (x - 2)(x - 4).

⇒ x² - 9 = (x - 3)(x + 3).

So:

$\Rightarrow \dfrac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)} = \dfrac{(x + 3)(x - 2)(x - 3)(x - 4)}{(x - 2)(x - 4)(x - 3)(x + 3)} \\[1em] = 1.$

Hence, $\dfrac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)} = 1$.

Simplify the following rational expression assuming that the expression in the denominator is not equal to zero:

$\dfrac{p^4 - 16}{p^2 - 4p + 4}$

Answer

Factoring the numerator:

⇒ p⁴ - 16 = (p²)² - (4)² = (p² - 4)(p² + 4)

= (p - 2)(p + 2)(p² + 4).

Factoring the denominator:

⇒ p² - 4p + 4 = (p - 2)².

So:

$\Rightarrow \dfrac{p^4 - 16}{p^2 - 4p + 4} = \dfrac{(p - 2)(p + 2)(p^2 + 4)}{(p - 2)^2} \\[1em] = \dfrac{(p + 2)(p^2 + 4)}{p - 2}.$

Hence, $\dfrac{p^4 - 16}{p^2 - 4p + 4} = \dfrac{(p + 2)(p^2 + 4)}{p - 2}$.

Use suitable identities to find the following products:

(–3x + 4)²

Answer

Using the identity (a + b)² = a² + 2ab + b², with a = –3x and b = 4:

⇒ (-3x + 4)² = (-3x)² + 2(-3x)(4) + (4)²

= 9x² - 24x + 16.

Hence, (–3x + 4)² = 9x² – 24x + 16.

Use suitable identities to find the following products:

(2s + 7) (2s – 7)

Answer

Using the identity (a + b)(a – b) = a² – b², with a = 2s and b = 7:

⇒ (2s + 7)(2s - 7) = (2s)² - (7)²

= 4s² - 49.

Hence, (2s + 7)(2s – 7) = 4s² – 49.

Use suitable identities to find the following products:

$\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right)$

Answer

Using the identity (a + b)(a – b) = a² – b², with a = p² and $b = \dfrac{1}{2}$:

$\Rightarrow \left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right) = (p^2)^2 - \left(\dfrac{1}{2}\right)^2 \\[1em] = p^4 - \dfrac{1}{4}.$

Hence, $\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right) = p^4 - \dfrac{1}{4}$.

Use suitable identities to find the following products:

(2n + 7) (2n – 7)

Answer

Using the identity (a + b)(a – b) = a² – b², with a = 2n and b = 7:

⇒ (2n + 7)(2n - 7) = (2n)² - (7)²

= 4n² - 49.

Hence, (2n + 7)(2n – 7) = 4n² – 49.

Use suitable identities to find the following products:

(s – 2t) (s² + 2st + 4t²)

Answer

Using the identity (a – b)(a² + ab + b²) = a³ – b³, with a = s and b = 2t:

⇒ (s - 2t)(s² + 2st + 4t²) = (s - 2t)[s² + s(2t) + (2t)²]

= s³ - (2t)³

= s³ - 8t³.

Hence, (s – 2t)(s² + 2st + 4t²) = s³ – 8t³.

Use suitable identities to find the following products:

$\left(\dfrac{1}{2r} - 4r\right)^2$

Answer

Using the identity (a – b)² = a² – 2ab + b², with $a = \dfrac{1}{2r}$ and b = 4r:

$\Rightarrow \left(\dfrac{1}{2r} - 4r\right)^2 = \left(\dfrac{1}{2r}\right)^2 - 2\left(\dfrac{1}{2r}\right)(4r) + (4r)^2 \\[1em] = \dfrac{1}{4r^2} - 4 + 16r^2.$

Hence, $\left(\dfrac{1}{2r} - 4r\right)^2 = \dfrac{1}{4r^2} - 4 + 16r^2$.

Use suitable identities to find the following products:

(–3m + 4k – l)²

Answer

Using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca, with a = –3m, b = 4k, c = –l:

⇒ (-3m + 4k - l)² = (-3m)² + (4k)² + (-l)² + 2(-3m)(4k) + 2(4k)(-l) + 2(-3m)(-l)

= 9m² + 16k² + l² - 24mk - 8kl + 6ml.

Hence, (–3m + 4k – l)² = 9m² + 16k² + l² – 24mk – 8kl + 6ml.

Use suitable identities to find the following products:

$\left(x - \dfrac{1}{3}y\right)^3$

Answer

Using the identity (a – b)³ = a³ – 3a²b + 3ab² – b³, with a = x and $b = \dfrac{1}{3}y$:

$\Rightarrow \left(x - \dfrac{1}{3}y\right)^3 = x^3 - 3(x)^2\left(\dfrac{1}{3}y\right) + 3(x)\left(\dfrac{1}{3}y\right)^2 - \left(\dfrac{1}{3}y\right)^3 \\[1em] = x^3 - x^2y + \dfrac{xy^2}{3} - \dfrac{y^3}{27}.$

Hence, $\left(x - \dfrac{1}{3}y\right)^3 = x^3 - x^2y + \dfrac{xy^2}{3} - \dfrac{y^3}{27}$.

Use suitable identities to find the following products:

$\left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3$

Answer

Using the identity (a – b)³ = a³ – 3a²b + 3ab² – b³, with $a = \dfrac{7}{2}k$ and $b = \dfrac{2}{3}m$:

$\Rightarrow \left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3 = \left(\dfrac{7}{2}k\right)^3 - 3\left(\dfrac{7}{2}k\right)^2\left(\dfrac{2}{3}m\right) + 3\left(\dfrac{7}{2}k\right)\left(\dfrac{2}{3}m\right)^2 - \left(\dfrac{2}{3}m\right)^3 \\[1em] = \dfrac{343}{8}k^3 - 3 \times \dfrac{49}{4}k^2 \times \dfrac{2}{3}m + 3 \times \dfrac{7}{2}k \times \dfrac{4}{9}m^2 - \dfrac{8}{27}m^3 \\[1em] = \dfrac{343}{8}k^3 - \dfrac{49}{2}k^2m + \dfrac{14}{3}km^2 - \dfrac{8}{27}m^3.$

Hence, $\left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3 = \dfrac{343}{8}k^3 - \dfrac{49}{2}k^2m + \dfrac{14}{3}km^2 - \dfrac{8}{27}m^3$.

Find the values using suitable identities:

17 × 21

Answer

Writing 17 as (19 – 2) and 21 as (19 + 2), and using (a + b)(a – b) = a² – b²:

⇒ 17 × 21 = (19 - 2)(19 + 2) = (19)² - (2)²

= 361 - 4

= 357.

Hence, 17 × 21 = 357.

Find the values using suitable identities:

104 × 96

Answer

Writing 104 as (100 + 4) and 96 as (100 – 4), and using (a + b)(a – b) = a² – b²:

⇒ 104 × 96 = (100 + 4)(100 - 4) = (100)² - (4)²

= 10000 - 16

= 9984.

Hence, 104 × 96 = 9984.

Find the values using suitable identities:

24 × 16

Answer

Writing 24 as (20 + 4) and 16 as (20 – 4), and using (a + b)(a – b) = a² – b²:

⇒ 24 × 16 = (20 + 4)(20 - 4) = (20)² - (4)²

= 400 - 16

= 384.

Hence, 24 × 16 = 384.

Find the values using suitable identities:

147³

Answer

Writing 147 as (150 – 3) and using (a – b)³ = a³ – 3a²b + 3ab² – b³:

⇒ (147)³ = (150 - 3)³

= (150)³ - 3(150)²(3) + 3(150)(3)² - (3)³

= 3375000 - 202500 + 4050 - 27

= 3176523.

Hence, (147)³ = 3176523.

Find the values using suitable identities:

199³

Answer

Writing 199 as (200 – 1) and using (a – b)³ = a³ – 3a²b + 3ab² – b³:

⇒ (199)³ = (200 - 1)³

= (200)³ - 3(200)²(1) + 3(200)(1)² - (1)³

= 8000000 - 120000 + 600 - 1

= 7880599.

Hence, (199)³ = 7880599.

Find the values using suitable identities:

127³

Answer

Writing 127 as (130 – 3) and using (a – b)³ = a³ – 3a²b + 3ab² – b³:

⇒ (127)³ = (130 - 3)³

= (130)³ - 3(130)²(3) + 3(130)(3)² - (3)³

= 2197000 - 152100 + 3510 - 27

= 2048383.

Hence, (127)³ = 2048383.

Find the values using suitable identities:

(–107)³

Answer

Writing 107 as (100 + 7) and using (a + b)³ = a³ + 3a²b + 3ab² + b³:

⇒ (107)³ = (100 + 7)³

= (100)³ + 3(100)²(7) + 3(100)(7)² + (7)³

= 1000000 + 210000 + 14700 + 343

= 1225043.

So:

⇒ (-107)³ = -(107)³ = -1225043.

Hence, (–107)³ = –1225043.

Find the values using suitable identities:

(–299)³

Answer

Writing 299 as (300 – 1) and using (a – b)³ = a³ – 3a²b + 3ab² – b³:

⇒ (299)³ = (300 - 1)³

= (300)³ - 3(300)²(1) + 3(300)(1)² - (1)³

= 27000000 - 270000 + 900 - 1

= 26730899.

So:

⇒ (-299)³ = -(299)³ = -26730899.

Hence, (–299)³ = –26730899.

Factor the following algebraic expressions:

$4y^2 + 1 + \dfrac{1}{16y^2}$

Answer

Rewriting the expression in the form a² + 2ab + b²:

$\Rightarrow 4y^2 + 1 + \dfrac{1}{16y^2} = (2y)^2 + 2(2y)\left(\dfrac{1}{4y}\right) + \left(\dfrac{1}{4y}\right)^2 \\[1em] = \left(2y + \dfrac{1}{4y}\right)^2.$

Hence, $4y^2 + 1 + \dfrac{1}{16y^2} = \left(2y + \dfrac{1}{4y}\right)^2$.

Factor the following algebraic expressions:

$9m^2 - \dfrac{1}{25n^2}$

Answer

Using the identity a² – b² = (a + b)(a – b):

$\Rightarrow 9m^2 - \dfrac{1}{25n^2} = (3m)^2 - \left(\dfrac{1}{5n}\right)^2 \\[1em] = \left(3m - \dfrac{1}{5n}\right)\left(3m + \dfrac{1}{5n}\right).$

Hence, $9m^2 - \dfrac{1}{25n^2} = \left(3m - \dfrac{1}{5n}\right)\left(3m + \dfrac{1}{5n}\right)$.

Factor the following algebraic expressions:

$27b^3 - \dfrac{1}{64b^3}$

Answer

Using the identity a³ – b³ = (a – b)(a² + ab + b²), with a = 3b and $b = \dfrac{1}{4b}$:

$\Rightarrow 27b^3 - \dfrac{1}{64b^3} = (3b)^3 - \left(\dfrac{1}{4b}\right)^3 \\[1em] = \left(3b - \dfrac{1}{4b}\right)\left[(3b)^2 + (3b)\left(\dfrac{1}{4b}\right) + \left(\dfrac{1}{4b}\right)^2\right] \\[1em] = \left(3b - \dfrac{1}{4b}\right)\left(9b^2 + \dfrac{3}{4} + \dfrac{1}{16b^2}\right).$

Hence, $27b^3 - \dfrac{1}{64b^3} = \left(3b - \dfrac{1}{4b}\right)\left(9b^2 + \dfrac{3}{4} + \dfrac{1}{16b^2}\right)$.

Factor the following algebraic expressions:

$x^2 + \dfrac{5x}{6} + \dfrac{1}{6}$

Answer

We need two numbers a and b such that $a + b = \dfrac{5}{6}$ and $ab = \dfrac{1}{6}$.

Choosing $a = \dfrac{1}{2}$ and $b = \dfrac{1}{3}$: $a + b = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$ and $ab = \dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$ .

$\Rightarrow x^2 + \dfrac{5x}{6} + \dfrac{1}{6} = x^2 + \dfrac{x}{2} + \dfrac{x}{3} + \dfrac{1}{6} \\[1em] = x\left(x + \dfrac{1}{2}\right) + \dfrac{1}{3}\left(x + \dfrac{1}{2}\right) \\[1em] = \left(x + \dfrac{1}{2}\right)\left(x + \dfrac{1}{3}\right).$

Hence, $x^2 + \dfrac{5x}{6} + \dfrac{1}{6} = \left(x + \dfrac{1}{2}\right)\left(x + \dfrac{1}{3}\right)$.

Factor the following algebraic expressions:

$27u^3 - \dfrac{1}{125} - \dfrac{27u^2}{5} + \dfrac{9u}{25}$

Answer

Rewriting the expression in the form a³ – 3a²b + 3ab² – b³, with a = 3u and $b = \dfrac{1}{5}$:

$\Rightarrow 27u^3 - \dfrac{27u^2}{5} + \dfrac{9u}{25} - \dfrac{1}{125} \\[1em] = (3u)^3 - 3(3u)^2\left(\dfrac{1}{5}\right) + 3(3u)\left(\dfrac{1}{5}\right)^2 - \left(\dfrac{1}{5}\right)^3 \\[1em] = \left(3u - \dfrac{1}{5}\right)^3.$

Hence, $27u^3 - \dfrac{1}{125} - \dfrac{27u^2}{5} + \dfrac{9u}{25} = \left(3u - \dfrac{1}{5}\right)^3$.

Factor the following algebraic expressions:

$64y^3 + \dfrac{1}{125}z^3$

Answer

Using the identity a³ + b³ = (a + b)(a² – ab + b²), with a = 4y and $b = \dfrac{z}{5}$:

$\Rightarrow 64y^3 + \dfrac{1}{125}z^3 = (4y)^3 + \left(\dfrac{z}{5}\right)^3 \\[1em] = \left(4y + \dfrac{z}{5}\right)\left[(4y)^2 - (4y)\left(\dfrac{z}{5}\right) + \left(\dfrac{z}{5}\right)^2\right] \\[1em] = \left(4y + \dfrac{z}{5}\right)\left(16y^2 - \dfrac{4yz}{5} + \dfrac{z^2}{25}\right).$

Hence, $64y^3 + \dfrac{1}{125}z^3 = \left(4y + \dfrac{z}{5}\right)\left(16y^2 - \dfrac{4yz}{5} + \dfrac{z^2}{25}\right)$.

Factor the following algebraic expressions:

p³ + 27q³ + r³ – 9pqr

Answer

Using the identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca), with a = p, b = 3q, c = r:

⇒ p³ + 27q³ + r³ - 9pqr = p³ + (3q)³ + r³ - 3(p)(3q)(r)

= (p + 3q + r)[p² + (3q)² + r² - p(3q) - (3q)(r) - p(r)]

= (p + 3q + r)(p² + 9q² + r² - 3pq - 3qr - pr).

Hence, p³ + 27q³ + r³ – 9pqr = (p + 3q + r)(p² + 9q² + r² – 3pq – 3qr – pr).

Factor the following algebraic expressions:

9m² – 12m + 4

Answer

Rewriting the expression in the form a² – 2ab + b²:

⇒ 9m² - 12m + 4 = (3m)² - 2(3m)(2) + (2)²

= (3m - 2)².

Hence, 9m² – 12m + 4 = (3m – 2)².

Factor the following algebraic expressions:

$9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz$

Answer

Multiplying and dividing by 3:

$\Rightarrow 9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz \\[1em] = \dfrac{1}{3}\left(27x^3 - 8y^3 + z^3 + 18xyz\right).$

Now rewriting using the identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca), with a = 3x, b = –2y, c = z:

⇒ 27x³ - 8y³ + z³ + 18xyz = (3x)³ + (-2y)³ + z³ - 3(3x)(-2y)(z)

= (3x - 2y + z)[(3x)² + (-2y)² + z² - (3x)(-2y) - (-2y)(z) - (3x)(z)]

= (3x - 2y + z)(9x² + 4y² + z² + 6xy + 2yz - 3xz).

So:

$\Rightarrow 9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz = \dfrac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3xz).$

Hence, $9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz = \dfrac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3xz)$.

Factor the following algebraic expressions:

4x² + 9y² + 36z² + 12xy + 36yz + 24xz

Answer

Rewriting the expression in the form a² + b² + c² + 2ab + 2bc + 2ca, with a = 2x, b = 3y, c = 6z:

⇒ 4x² + 9y² + 36z² + 12xy + 36yz + 24xz

= (2x)² + (3y)² + (6z)² + 2(2x)(3y) + 2(3y)(6z) + 2(2x)(6z)

= (2x + 3y + 6z)².

Hence, 4x² + 9y² + 36z² + 12xz + 36yz + 24xy = (2x + 3y + 6z)².

Factor the following algebraic expressions:

$27u^3 - \dfrac{1}{216} - \dfrac{9u^2}{2} + \dfrac{u}{4}$