Expansions — Exercise 4(B)

If x² - 2x + 1 = 0; the value of $x^4 + \dfrac{1}{x^4}$ is equal to :

1. 9

2. 7

3. 3

4. 2

Answer

Given,

⇒ x² - 2x + 1 = 0

⇒ x² + 1 = 2x

Dividing above equation by x, we get :

⇒ $\dfrac{x^2 + 1}{x} = \dfrac{2x}{x}$

⇒ $x + \dfrac{1}{x}$ = 2.

Squaring both sides we get :

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 2^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x} = 4 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 = 4 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2.$

Squaring both sides we get :

$\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = 2^2 \\[1em] \Rightarrow (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} = 4 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} + 2 = 4 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 4 - 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 2.$

Hence, Option 4 is the correct option.

If a = 5, b = -2 and c = -3, a³ + b³ + c³ is equal to :

1. -90

2. 90

3. 60

4. -60

Answer

By property,

If a + b + c = 0, then

a³ + b³ + c³ = 3abc

Since, 5 + (-2) + (-3) = 5 - 2 - 3 = 0.

∴ a³ + b³ + c³ = 3 × 5 × -2 × -3 = 90.

Hence, Option 2 is the correct option.

If $x^2 + \dfrac{1}{x^2} = 2$, the value of $x^2 - \dfrac{1}{x^2}$ is :

1. 0

2. 2

3. $\pm 2$

4. 8

Answer

By formula,

$\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - \Big(x^2 - \dfrac{1}{x^2}\Big)^2$ = 4.

Substituting values we get :

$\Rightarrow 2^2 - \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4 \\[1em] \Rightarrow 4 - \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4 - 4 \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 0 \\[1em] \Rightarrow x^2 - \dfrac{1}{x^2} = 0.$

Hence, Option 1 is the correct option.

If x + 3y + 2z = 0, the value of x³ + 27y³ + 8z³ is :

1. 9xyz

2. 6xyz

3. 18xyz

4. xyz

Answer

By property,

If a + b + c = 0, then

a³ + b³ + c³ = 3abc ....(1)

Comparing equation x + 3y + 2z = 0, with a + b + c = 0, we get :

a = x, b = 3y and c = 2z.

Substituting values in equation (1), we get :

x³ + (3y)³ + (2z)³ = 3 × x × 3y × 2z = 18xyz.

Hence, Option 3 is the correct option.

Find the cube of 3a - 2b.

Answer

Solving,

(3a - 2b)³ = (3a)³ - (2b)³ - 3 × 3a × 2b(3a - 2b)

= 27a³ - 8b³ - 18ab(3a - 2b)

= 27a³ - 8b³ - 54a²b + 36ab².

Hence, cube of 3a - 2b = 27a³ - 8b³ - 54a²b + 36ab².

Find the cube of 5a + 3b.

Answer

Solving,

(5a + 3b)³ = (5a)³ + (3b)³ + 3 × 5a × 3b(5a + 3b)

= 125a³ + 27b³ + 45ab(5a + 3b)

= 125a³ + 27b³ + 225a²b + 135ab².

Hence, cube of 5a + 3b = 125a³ + 27b³ + 225a²b + 135ab².

Find the cube of 2a + $\dfrac{1}{2a}$, (a ≠ 0)

Answer

Solving,

$\Big(2a + \dfrac{1}{2a}\Big)^3 = (2a)^3 + \Big(\dfrac{1}{2a}\Big)^3 + 3 \times 2a \times \dfrac{1}{2a} \times \Big(2a + \dfrac{1}{2a}\Big) \\[1em] \Rightarrow 8a^3 + \dfrac{1}{8a^3} + 3\Big(2a + \dfrac{1}{2a}\Big) \\[1em] \Rightarrow 8a^3 + \dfrac{1}{8a^3} + 6a + \dfrac{3}{2a}.$

Hence, $\Big(2a + \dfrac{1}{2a}\Big)^3 = 8a^3 + \dfrac{1}{8a^3} + 6a + \dfrac{3}{2a}$.

Find the cube of $3a - \dfrac{1}{a}$ (a ≠ 0)

Answer

Solving,

$\Rightarrow \Big(3a - \dfrac{1}{a}\Big)^3 = (3a)^3 - \Big(\dfrac{1}{a}\Big)^3 - 3 \times 3a \times \dfrac{1}{a} \times \Big(3a - \dfrac{1}{a}\Big) \\[1em] = 27a^3 - \dfrac{1}{a^3} - 9\Big(3a - \dfrac{1}{a}\Big) \\[1em] = 27a^3 - \dfrac{1}{a^3} - 27a + \dfrac{9}{a}.$

Hence, $\Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - \dfrac{1}{a^3} - 27a + \dfrac{9}{a}.$

If $a^2 + \dfrac{1}{a^2} = 47$ and a ≠ 0; find :

(i) $a + \dfrac{1}{a}$

(ii) $a^3 + \dfrac{1}{a^3}$

Answer

(i) By formula,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 47 + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 49 \\[1em] \Rightarrow a + \dfrac{1}{a} = \sqrt{49} \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 7.$

Hence, $a + \dfrac{1}{a} = \pm 7.$

(ii) By formula,

$\Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)$ ........(1)

Substituting $a + \dfrac{1}{a} = 7$ in equation (1), we get :

$\Rightarrow 7^3 = a^3 + \dfrac{1}{a^3} + 3 \times 7 \\[1em] \Rightarrow 343 = a^3 + \dfrac{1}{a^3} + 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 343 - 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 322.$

Substituting $a + \dfrac{1}{a} = -7$ in equation (1), we get :

$\Rightarrow (-7)^3 = a^3 + \dfrac{1}{a^3} + 3 \times -7 \\[1em] \Rightarrow -343 = a^3 + \dfrac{1}{a^3} - 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -343 + 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -322.$

Hence, $a^3 + \dfrac{1}{a^3} = \pm 322.$

If $a^2 + \dfrac{1}{a^2} = 18$ and a ≠ 0; find :

(i) $a - \dfrac{1}{a}$

(ii) $a^3 - \dfrac{1}{a^3}$

Answer

(i) By formula,

$\Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 18 - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 16 \\[1em] \Rightarrow a - \dfrac{1}{a} = \sqrt{16} \\[1em] \Rightarrow a - \dfrac{1}{a} = \pm 4.$

Hence, $a - \dfrac{1}{a} = \pm 4$.

(ii) By formula,

$\Big(a - \dfrac{1}{a}\Big)^3 = a^3 - \dfrac{1}{a^3} -3\Big(a - \dfrac{1}{a}\Big)$

Substituting $a - \dfrac{1}{a} = 4$ we get :

$\Rightarrow 4^3 = a^3 - \dfrac{1}{a^3} - 3 \times 4 \\[1em] \Rightarrow 64 = a^3 - \dfrac{1}{a^3} - 12 \\[1em] \Rightarrow 64 + 12 = a^3 - \dfrac{1}{a^3} \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} = 76.$

Substituting $a - \dfrac{1}{a} = -4$ we get :

$\Rightarrow (-4)^3 = a^3 - \dfrac{1}{a^3} - 3 \times -4 \\[1em] \Rightarrow -64 = a^3 - \dfrac{1}{a^3} + 12 \\[1em] \Rightarrow -64 - 12 = a^3 - \dfrac{1}{a^3} \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} = -76.$

Hence, $a^3 - \dfrac{1}{a^3} = \pm 76$.

If $a + \dfrac{1}{a} = p$ and a ≠ 0; then show that :

$a^3 + \dfrac{1}{a^3} = p(p^2 - 3)$

Answer

By formula,

$a^3 + \dfrac{1}{a^3} = \Big(a + \dfrac{1}{a}\Big)^3 - 3\Big(a + \dfrac{1}{a}\Big) \\[1em] = p^3 - 3 \times p \\[1em] = p^3 - 3p \\[1em] = p(p^2 - 3)$

Hence, proved that $a^3 + \dfrac{1}{a^3} = p(p^2 - 3)$.

If a + 2b = 5; then show that :

a³ + 8b³ + 30ab = 125.

Answer

Given,

⇒ a + 2b = 5

Cubing both sides we get :

⇒ (a + 2b)³ = 5³

⇒ a³ + (2b)³ + 3 × a × 2b (a + 2b) = 125

⇒ a³ + 8b³ + 6ab(a + 2b) = 125

⇒ a³ + 8b³ + 6ab × 5 = 125

⇒ a³ + 8b³ + 30ab = 125.

Hence, proved that a³ + 8b³ + 30ab = 125.

If $\Big(a + \dfrac{1}{a}\Big)^2 = 3$ and a ≠ 0; then show that :

$a^3 + \dfrac{1}{a^3} = 0.$

Answer

Given,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 3 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm\sqrt{3}$

By formula,

$\Rightarrow a^3 + \dfrac{1}{a^3} = \Big(a + \dfrac{1}{a}\Big)^3 - 3\Big(a + \dfrac{1}{a}\Big)$

Substituting $a + \dfrac{1}{a} = -\sqrt{3}$, we get :

$\Rightarrow a^3 + \dfrac{1}{a^3} = (-\sqrt{3})^3 - 3 \times -\sqrt{3} \\[1em] = -3\sqrt{3} + 3\sqrt{3} \\[1em] = 0.$

Substituting $a + \dfrac{1}{a} = \sqrt{3}$, we get :

$\Rightarrow a^3 + \dfrac{1}{a^3} = (\sqrt{3})^3 - 3 \times \sqrt{3} \\[1em] = 3\sqrt{3} - 3\sqrt{3} \\[1em] = 0.$

Hence, proved that $a^3 + \dfrac{1}{a^3} = 0.$

If a + 2b + c = 0; then show that :

a³ + 8b³ + c³ = 6abc.

Answer

Given,

a + 2b + c = 0 ........(1)

By property,

If x + y + z = 0, then

⇒ x³ + y³ + z³ = 3xyz .........(2)

Comparing,

Eq. 1 with x + y + z = 0, we get :

x = a, y = 2b and z = c.

Then,

Substituting value of x, y and z in Eq. 2, we get :

⇒ a³ + (2b)³ + c³ = 3 × a × 2b × c

⇒ a³ + 8b³ + c³ = 6abc.

Hence, proved that a³ + 8b³ + c³ = 6abc.

Use property to evaluate:

(i) 9³ - 5³ - 4³

(ii) 38³ + (-26)³ + (-12)³

Answer

(i) By property,

If a + b + c = 0, then

⇒ a³ + b³ + c³ = 3abc

Given,

⇒ 9³ - 5³ - 4³

⇒ 9³ + (-5)³ + (-4)³

Let a = 9, b = -5 and c = -4.

∴ a + b + c = 9 + (-5) + (-4) = 9 - 5 - 4 = 0

∴ 9³ + (-5)³ + (-4)³ = 3 × 9 × (-5) × (-4) = 540.

Hence, 9³ - 5³ - 4³ = 540.

(ii) By property,

If a + b + c = 0, then

⇒ a³ + b³ + c³ = 3abc

Given,

⇒ 38³ + (-26)³ + (-12)³

Let a = 38, b = -26 and c = -12.

∴ a + b + c = 38 + (-26) + (-12) = 38 - 26 - 12 = 0

∴ 38³ + (-26)³ + (-12)³ = 3 × 38 × (-26) × (-12) = 35,568.

Hence, 38³ + (-26)³ + (-12)³ = 35568.

If a≠ 0 and $a - \dfrac{1}{a} = 3$; find :

(i) $a^2 + \dfrac{1}{a^2}$

(ii) $a^3 - \dfrac{1}{a^3}$

Answer

(i) By formula,

$\Rightarrow a^2 + \dfrac{1}{a^2} = \Big(a - \dfrac{1}{a}\Big)^2 + 2 \\[1em] = 3^2 + 2 \\[1em] = 9 + 2 = 11.$

Hence, $a^2 + \dfrac{1}{a^2} = 11$.

(ii) By formula,

$\Rightarrow \Big(a - \dfrac{1}{a}\Big)^3 = a^3 - \dfrac{1}{a^3} - 3\Big(a -\dfrac{1}{a}\Big) \\[1em] \Rightarrow 3^3 = a^3 - \dfrac{1}{a^3} - 3 \times 3 \\[1em] \Rightarrow 27 = a^3 - \dfrac{1}{a^3} - 9 \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} = 36.$

Hence, $a^3 - \dfrac{1}{a^3} = 36$.

If a ≠ 0 and $a - \dfrac{1}{a} = 4$; find :

(i) $a^2 + \dfrac{1}{a^2}$

(ii) $a^4 + \dfrac{1}{a^4}$

(iii) $a^3 - \dfrac{1}{a^3}$

Answer

(i) By formula,

$\Rightarrow a^2 + \dfrac{1}{a^2} = \Big(a - \dfrac{1}{a}\Big)^2 + 2$

Substituting values we get :

$\Rightarrow a^2 + \dfrac{1}{a^2} = 4^2 + 2 \\[1em] = 16 + 2 \\[1em] = 18.$

Hence, $a^2 + \dfrac{1}{a^2} = 18$.

(ii) By formula,

$\Rightarrow a^4 + \dfrac{1}{a^4} = \Big(a^2 + \dfrac{1}{a^2}\Big)^2 - 2$

Substituting values we get :

$\Rightarrow a^4 + \dfrac{1}{a^4} = 18^2 - 2 \\[1em] = 324 - 2 \\[1em] = 322.$

Hence, $a^4 + \dfrac{1}{a^4} = 322$.

(iii) By formula,

$\Rightarrow a^3 - \dfrac{1}{a^3} = \Big(a - \dfrac{1}{a}\Big)^3 + 3\Big(a - \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow a^3 - \dfrac{1}{a^3} = 4^3 + 3 \times 4 \\[1em] = 64 + 12 \\[1em] = 76.$

Hence, $a^3 - \dfrac{1}{a^3} = 76$.

If x ≠ 0 and $x + \dfrac{1}{x} = 2$; then show that :

$x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}$.

Answer

By formula,

$\Rightarrow x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2$

Substituting values we get :

$\Rightarrow x^2 + \dfrac{1}{x^2} = 2^2 - 2 \\[1em] = 4 - 2 \\[1em] = 2.$

By formula,

$\Rightarrow x^3 + \dfrac{1}{x^3} = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big)$

Substituting values we get :

$\Rightarrow x^3 + \dfrac{1}{x^3} = 2^3 - 3 \times 2 \\[1em] = 8 - 6 \\[1em] = 2.$

By formula,

$\Rightarrow x^4 + \dfrac{1}{x^4} = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2$

Substituting values we get :

$\Rightarrow x^4 + \dfrac{1}{x^4} = 2^2 - 2 \\[1em] = 4 - 2 \\[1em] = 2.$

Hence, proved that $x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}$.

If 2x - 3y = 10 and xy = 16, find the value of 8x³ - 27y³.

Answer

Given,

2x - 3y = 10

Cubing both sides we get :

⇒ (2x - 3y)³ = 10³

⇒ (2x)³ - (3y)³ - 3 × 2x × 3y(2x - 3y) = 1000

⇒ 8x³ - 27y³ - 18xy(2x - 3y) = 1000

⇒ 8x³ - 27y³ - 18 × 16 × 10 = 1000

⇒ 8x³ - 27y³ - 2880 = 1000

⇒ 8x³ - 27y³ = 3880.

Hence, 8x³ - 27y³ = 3880.

Expand (3x + 5y + 2z)(3x - 5y + 2z)

Answer

Given,

⇒ (3x + 5y + 2z)(3x - 5y + 2z)

⇒ 9x² - 15xy + 6xz + 15xy - 25y² + 10yz + 6zx - 10yz + 4z²

⇒ 9x² - 25y² + 4z² + 12xz.

Hence, (3x + 5y + 2z)(3x - 5y + 2z) = 9x² - 25y² + 4z² + 12xz.

Expand (3x - 5y - 2z)(3x - 5y + 2z)

Answer

Given,

$⇒ (3x - 5y - 2z)(3x - 5y + 2z) \\[1em] ⇒ 9x^2 - 15xy + 6zx - 15xy + 25y^2 - 10yz - 6zx + 10yz - 4z^2 \\[1em] ⇒ 9x^2 - 15xy + \cancel{6zx} - 15xy + 25y^2 - \cancel{10yz} - \cancel{6zx} + \cancel{10yz} - 4z^2 \\[1em] ⇒ 9x^2 - 30xy + 25y^2 - 4z^2 .$

Hence, (3x - 5y - 2z)(3x - 5y + 2z) = 9x² - 30xy + 25y² - 4z².

The sum of two numbers is 9 and their product is 20. Find the sum of their :

(i) squares

(ii) cubes.

Answer

Let two numbers be a and b.

Given,

Sum of numbers = 9 and Product = 20.

∴ a + b = 9 and ab = 20.

(i) Sum of squares = a² + b²

By formula,

⇒ (a + b)² = a² + b² + 2ab

Substituting values we get :

⇒ 9² = a² + b² + 2 × 20

⇒ 81 = a² + b² + 40

⇒ a² + b² = 81 - 40 = 41.

Hence, sum of squares = 41.

(ii) Sum of cubes = a³ + b³

By formula,

⇒ (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ 9³ = a³ + b³ + 3 × 20 × 9

⇒ 729 = a³ + b³ + 540

⇒ a³ + b³ = 729 - 540

⇒ a³ + b³ = 189.

Hence, a³ + b³ = 189.

Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find :

(i) sum of these numbers.

(ii) difference of their cubes.

(iii) sum of their cubes.

Answer

Given,

Difference of numbers = 5 and product = 24.

Let the two numbers be x and y.

∴ x - y = 5 and xy = 24.

(i) By formula,

(x + y)² = (x - y)² + 4xy

Substituting values we get :

⇒ (x + y)² = 5² + 4 × 24

⇒ (x + y)² = 25 + 96

⇒ (x + y)² = 121

⇒ (x + y) = $\sqrt{121} = \pm 11$.

Since, numbers are positive so sum cannot be negative.

Hence, sum of these numbers = 11.

(ii) By formula,

⇒ x³ - y³ = (x - y)³ + 3xy(x - y)

⇒ x³ - y³ = 5³ + 3 × 24 × 5

⇒ x³ - y³ = 125 + 360

⇒ x³ - y³ = 485.

Hence, difference of cubes of numbers = 485.

(iii) By formula,

⇒ x³ + y³ = (x + y)³ - 3xy(x + y)

Substituting x + y = 11, we get :

⇒ x³ + y³ = 11³ - 3 × 24 × 11

⇒ x³ + y³ = 1331 - 792

⇒ x³ + y³ = 539.

Hence, sum of cubes of numbers = 539.

If 4x² + y² = a and xy = b, find the value of 2x + y.

Answer

⇒ (2x + y)² = (2x)² + y² + 2 × 2x × y

⇒ (2x + y)² = 4x² + y² + 4xy

Substituting values we get,

⇒ (2x + y)² = a + 4b

⇒ (2x + y) = $\pm \sqrt{a + 4b}$.

Hence, (2x + y) = $\pm \sqrt{a + 4b}$.