Expansions — Chapter Test

Find the expansions of the following :

(i) (2x + 3y + 5)(2x + 3y - 5)

(ii) (6 - 4a - 7b)²

(iii) (7 - 3xy)³

(iv) (x + y + 2)³

Answer

(i) On solving,

⇒ (2x + 3y + 5)(2x + 3y - 5) = (2x + 3y)² - 5²

⇒ (2x + 3y + 5)(2x + 3y - 5) = 4x² + 9y² + 12xy - 25.

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

(ii) On solving,

⇒ (6 - 4a - 7b)² = (6 - 4a)² + (7b)² - 2(6 - 4a)(7b)

⇒ (6 - 4a - 7b)² = 36 + 16a² - 48a + 49b² - 14b(6 - 4a)

⇒ (6 - 4a - 7b)² = 36 + 16a² + 49b² - 48a + 56ab - 84b.

Hence, (6 - 4a - 7b)² = 36 + 16a² + 49b² - 48a + 56ab - 84b.

(iii) On solving,

⇒ (7 - 3xy)³ = (7)³ - (3xy)³ - 3(7)(3xy)(7 - 3xy)

⇒ (7 - 3xy)³ = 343 - 27x³y³ - 63xy(7 - 3xy)

⇒ (7 - 3xy)³ = 343 - 27x³y³ - 441xy + 189x²y².

Hence, (7 - 3xy)³ = 343 - 27x³y³ - 441xy + 189x²y².

(iv) On solving,

⇒ (x + y + 2)³ = (x)³ + (y + 2)³ + 3(x)(y + 2)(x + y + 2)

⇒ (x + y + 2)³ = x³ + y³ + 2³ + 3(y)(2)(y + 2) + (3xy + 6x)(x + y + 2)

⇒ (x + y + 2)³ = x³ + y³ + 8 + 6y(y + 2) + 3xy(x + y + 2) + 6x(x + y + 2)

⇒ (x + y + 2)³ = x³ + y³ + 8 + 6y² + 12y + 3x²y + 3xy² + 6xy + 6x² + 6xy + 12x

⇒ (x + y + 2)³ = x³ + y³ + 3x²y + 3xy² + 6x² + 6y² + 12xy + 12x + 12y + 8.

Hence, (x + y + 2)³ = x³ + y³ + 3x²y + 3xy² + 6x² + 6y² + 12xy + 12x + 12y + 8.

Simplify (x - 2)(x + 2)(x² + 4)(x⁴ + 16).

Answer

As (a - b)(a + b) = (a² - b²)

On solving,

$(x - 2)(x + 2)(x^2 + 4)(x^2 + 16) = (x^2 - 4)(x^2 + 4)(x^4 + 16) \\[1em] = (x^4 - 16)(x^4 + 16) \\[1em] = ((x^4)^2 - 16^2) \\[1em] = x^8 - 256.$

On simplifying we get, (x - 2)(x + 2)(x² + 4)(x⁴ + 16) = x⁸ - 256.

Evaluate 1002 × 998 by using a special product.

Answer

We can write 1002 × 998 as (1000 + 2)(1000 - 2).

As (a - b)(a + b) = (a² - b²)

We get,

(1000 + 2)(1000 - 2) = 1000² - 2² = 1000000 - 4 = 999996.

Hence, 1002 × 998 = 999996.

If a + 2b + 3c = 0, prove that a³ + 8b³ + 27c³ = 18abc.

Answer

Given,

a + 2b + 3c = 0
a + 2b = -3c.

Cubing both sides,

(a + 2b)³ = (-3c)³

a³ + (2b)³ + 3(a)(2b)(a + 2b) = -27c³

a³ + 8b³ + 6ab(-3c) = -27c³

a³ + 8b³ - 18abc = -27c³

a³ + 8b³ + 27c³ = 18abc.

Hence, proved that a³ + 8b³ + 27c³ = 18abc.

If 2x = 3y - 5, then find the value of 8x³ - 27y³ - 90xy + 125.

Answer

Given, 2x = 3y - 5 or 2x - 3y = -5

Cubing both sides we get,

⇒ (2x - 3y)³ = (-5)³

(2x)³ - (3y)³ - 3(2x)(3y)(2x - 3y) = -125

⇒ 8x³ - 27y³ - 18xy(2x - 3y) = -125

⇒ 8x³ - 27y³ - 18xy(-5) = -125

⇒ 8x³ - 27y³ + 90xy = -125

⇒ 8x³ - 27y³ + 90xy + 125 = 0.

Hence, 8x³ - 27y³ + 90xy + 125 = 0.

If $a^2 - \dfrac{1}{a^2} = 5, \text{ evaluate } a^4 + \dfrac{1}{a^4}$.

Answer

We know that,

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

Substituting values we get,

$a^4 + \dfrac{1}{a^4} = (5)^2 + 2 = 25 + 2 = 27.$

Hence, $a^4 + \dfrac{1}{a^4} = 27.$

If $a + \dfrac{1}{a} = p \text{ and } a - \dfrac{1}{a} = q$, find the relation between p and q.

Answer

We know that,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \space .....(i)\\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \space .....(ii)$

Subtracting eq. (ii) from (i) we get,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \text{Substituting values we get}, \\[1em] \Rightarrow p^2 - q^2 = 4.$

Hence, p² - q² = 4.

If $\dfrac{a^2 + 1}{a} = 4$, find the value of $2a^3 + \dfrac{2}{a^3}$.

Answer

Given,

$\phantom{\therefore} \dfrac{a^2 + 1}{a} = 4 \\[1em] \therefore a + \dfrac{1}{a} = 4.$

Solving,

$\Rightarrow 2a^3 + \dfrac{2}{a^3} = 2\Big(a^3 + \dfrac{1}{a^3}\Big) \\[1em] = 2\Big[\Big(a + \dfrac{1}{a}\Big)^3 - 3\Big(a + \dfrac{1}{a}\Big)\Big] \\[1em] = 2[4^3 - 3 \times 4] \\[1em] = 2[64 - 12] \\[1em] = 2 \times 52 \\[1em] = 104.$

Hence, the value of $2a^3 + \dfrac{2}{a^3}$ = 104.

If $x = \dfrac{1}{4 - x}$, find the values of

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

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

(iii) $x^6 + \dfrac{1}{x^6}$

Answer

(i) Given,

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

On dividing above equation by x,

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

Hence, the value of $x + \dfrac{1}{x}$ = 4.

(ii) We know that,

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

Substituting values we get,

$\Big(x^3 + \dfrac{1}{x^3}\Big) = 4^3 - 3 \times 4 = 64 - 12 = 52.$

Hence, the value of $x^3 + \dfrac{1}{x^3}$ = 52.

(iii) We know,

$\Big(x^6 + \dfrac{1}{x^6}\Big) = \Big(x^3 + \dfrac{1}{x^3}\Big)^2 - 2$.

Substituting values we get,

$\Big(x^6 + \dfrac{1}{x^6}\Big) = 52^2 - 2 = 2704 - 2 = 2702.$

Hence, the value of $x^6 + \dfrac{1}{x^6}$ = 2702.

If $x - \dfrac{1}{x} = 3 + 2\sqrt{2}$, find the value of $\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big)$.

Answer

We know that,

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

Substituting values we get,

$x^3 - \dfrac{1}{x^3} = (3 + 2\sqrt{2})^3 + 3(3 + 2\sqrt{2}) \\[1em] = (3)^3 + (2\sqrt{2})^3 + 3(3)(2\sqrt{2})(3 + 2\sqrt{2}) + 9 + 6\sqrt{2} \\[1em] = 27 + 16\sqrt{2} + 18\sqrt{2}(3 + 2\sqrt{2}) + 9 + 6\sqrt{2} \\[1em] = 27 + 9 + 16\sqrt{2} + 54\sqrt{2} + 72 + 6\sqrt{2} \\[1em] = 108 + 76\sqrt{2}.$

So, $\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big) = \dfrac{1}{4} \times (108 + 76\sqrt{2}) = 27 + 19\sqrt{2}$.

Hence, $\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big) = 27 + 19\sqrt{2}$.

If $x + \dfrac{1}{x} = 3\dfrac{1}{3}$, find the value of $x^3 - \dfrac{1}{x^3}$.

Answer

We know that,

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

Substituting values we get,

$x - \dfrac{1}{x} = \sqrt{\Big(3\dfrac{1}{3}\Big)^2 - 4} \\[1em] = \sqrt{\Big(\dfrac{10}{3}\Big)^2 - 4} \\[1em] = \sqrt{\dfrac{100}{9} - 4} \\[1em] = \sqrt{\dfrac{100 - 36}{9}} \\[1em] = \sqrt{\dfrac{64}{9}} \\[1em] = \pm \dfrac{8}{3}.$

We know that,

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

Substituting values we get,

$x^3 - \dfrac{1}{x^3} = \Big(\pm \dfrac{8}{3}\Big)^3 + 3 \times \pm \Big(\dfrac{8}{3}\Big) \\[1em] = \pm \dfrac{512}{27} \pm 8 \\[1em] = \dfrac{\pm 512 \pm 216}{27} \\[1em] = \pm \dfrac{728}{27} \\[1em] = \pm 26\dfrac{26}{27}.$

Hence, the value of $x^3 - \dfrac{1}{x^3} = \pm 26\dfrac{26}{27}$.

If $x = 2 - \sqrt{3}$ then find the value of $x^3 - \dfrac{1}{x^3}$.

Answer

Given,

$\Rightarrow x = 2 - \sqrt{3} \\[1em] \therefore \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{2 + \sqrt{3}}{4 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}.$

So,

$x - \dfrac{1}{x} = 2 - \sqrt{3} - (2 + \sqrt{3}) = -2\sqrt{3}.$

Cubing both sides we get,

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^3 = (-2\sqrt{3})^3 \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} - 3(x)\Big(\dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) = -24\sqrt{3} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} - 3 \times -2\sqrt{3} = -24\sqrt{3} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} + 6\sqrt{3} = -24\sqrt{3} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = -24\sqrt{3} - 6\sqrt{3} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = -30\sqrt{3}.$

Hence, $x^3 - \dfrac{1}{x^3} = -30\sqrt{3}$.

If the sum of two numbers is 7 and sum of their cubes is 133, find the sum of their squares.

Answer

Let the two numbers be x and y.

Given,

Sum of two numbers is 7.

∴ x + y = 7

Sum of cubes of two numbers is 133.

∴ x³ + y³ = 133

By formula,

⇒ (x + y)³ = x³ + y³ + 3xy(x + y)

Substituting values we get :

⇒ 7³ = 133 + 3xy × 7

⇒ 343 = 133 + 21xy

⇒ 21xy = 343 - 133

⇒ 21xy = 210

⇒ xy = $\dfrac{210}{21}$ = 10.

By formula,

⇒ (x + y)² = x² + y² + 2xy

Substituting values we get :

⇒ 7² = x² + y² + 2 × 10

⇒ 49 = x² + y² + 20

⇒ x² + y² = 49 - 20 = 29.

Hence, sum of the squares of the numbers = 29.

If a - b = 7 and a³ - b³ = 133, find

(i) ab

(ii) a² + b².

Answer

(i) Given,

a - b = 7, cubing both sides we get,

⇒ (a - b)³ = 7³

⇒ a³ - b³ - 3ab(a - b) = 343

⇒ 133 - 3ab(7) = 343

⇒ 133 - 21ab = 343

⇒ -21ab = 343 - 133

⇒ -21ab = 210

⇒ ab = -10.

Hence, ab = -10.

(ii) We know that,

a² + b² = (a - b)² + 2ab

Substituting values we get,

⇒ a² + b² = (7)² + 2(-10)

⇒ a² + b² = 49 - 20 = 29.

Hence, a² + b² = 29.

Find the coefficient of x² in the expansion of

(x² + x + 1)² + (x² - x + 1)².

Answer

The above equation can be written as,

$(x^2 + x + 1)^2 + (x^2 - x + 1)^2 = {(x^2 + 1) + x}^2 + {(x^2 + 1) - x}^2 \\[1em] = 2{(x^2 + 1)^2 + x^2} \\[1em] = 2{(x^2)^2 + 1 + 2x^2 + x^2} \\[1em] = 2{x^4 + 3x^2 + 1} \\[1em] = 2x^4 + 6x^2 + 2.$

Hence, coefficient of x² = 6.

If x² + y² + z² = xy + yz + zx, prove that x = y = z

Answer

Given,

x² + y² + z² = xy + yz + zx

⇒ x² + y² + z² - (xy + yz + zx) = 0

⇒ x² + y² + z² - xy - yz - zx = 0

Multiplying by 2 on both sides,

⇒ 2x² + 2y² + 2z² - 2xy - 2yz - 2zx = 0

⇒ (x² + y² - 2xy) + (y² + z² - 2yz) + (z² + x² - 2xz) = 0

⇒ (x - y)² + (y - z)² + (z - x)² = 0

⇒ x - y = 0, y - z = 0 and z - x = 0

⇒ x = y, y = z and z = x

⇒ x = y = z

Hence, proved that if x² + y² + z² = xy + yz + zx, then x = y = z