Expansions — Competency Focused Questions

The value of (−28)³ + (18)³ + (10)³ is

1. 15120

2. -15120

3. -5040

4. none of these

Answer

Given,

(−28)³ + (18)³ + (10)³

We know that,

If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Since, -28 + 18 + 10 = 0

⇒ (−28)³ + 18³ + 10³ = 3 × (-28) × 18 × 10

⇒ (−28)³ + 18³ + 10³ = -15120.

Hence, Option 2 is the correct option.

If x + y = −4, the value of x³ + y³ − 12xy + 64 is :

1. 1

2. -1

3. 4

4. 0

Answer

Given,

x + y = -4

We know that,

⇒ x³ + y³ = (x + y)³ - 3xy(x + y)

⇒ x³ + y³ = (-4)³ - 3xy(-4)

⇒ x³ + y³ = -64 + 12xy

⇒ x³ + y³ - 12xy + 64 = 0.

Hence, Option 4 is the correct option.

If a + b + c = 0, then $\dfrac{(a + b)^2}{ab} + \dfrac{(b + c)^2}{bc} + \dfrac{(c + a)^2}{ac}$ is equal to :

1. 0

2. 1

3. 3

4. abc

Answer

Given,

a + b + c = 0

⇒ a + b = -c

⇒ b + c = -a

⇒ c + a = -b

Substituting the above values in $\dfrac{(a + b)^2}{ab} + \dfrac{(b + c)^2}{bc} + \dfrac{(c + a)^2}{ac}$, we get :

$\Rightarrow \dfrac{(-c)^2}{ab} + \dfrac{(-a)^2}{bc} + \dfrac{(-b)^2}{ca} \\[1em] \Rightarrow \dfrac{c^2}{ab} + \dfrac{a^2}{bc} + \dfrac{b^2}{ca} \\[1em] \Rightarrow \dfrac{a^3 + b^3 + c^3}{abc} \text{ ......(1)}$

We know that,

If, a + b + c = 0 then a³ + b³ + c³ = 3abc

Substituting the value of a³ + b³ + c³ in (1), we get :

$\Rightarrow \dfrac{3abc}{abc} \\[1em] \Rightarrow 3.$

Hence, option 3 is correct option.

If $\dfrac{a}{b} + \dfrac{b}{a} = -1$ (a, b ≠ 0), then the value of a³ - b³ is :

1. $\dfrac{1}{2}$

2. 1

3. -1

4. 0

Answer

Given,

$\Rightarrow \dfrac{a}{b} + \dfrac{b}{a} = -1 \\[1em] \Rightarrow \dfrac{a^2 + b^2}{ab} = -1 \\[1em] \Rightarrow a^2 + b^2 = -ab \\[1em] \Rightarrow a^2 + b^2 + ab = 0$

Using identity,

⇒ a³ - b³ = (a - b)(a² + b² + ab)

⇒ a³ - b³ = (a - b)(0)

⇒ a³ - b³ = 0.

Hence, option 4 is correct option.

If $x^4 + \dfrac{1}{x^4} = 119$, x > 1, then find the value of $x^3 - \dfrac{1}{x^3}$

Answer

Given,

$x^4 + \dfrac{1}{x^4} = 119$, and x > 1.

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

Using identity

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

Substitute the given value:

$\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = 119 + 2 \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = 121 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \sqrt{121} \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \pm 11$

Since $x \gt 1, x^2 + \dfrac{1}{x^2}$ must be positive.

So, $x^2 + \dfrac{1}{x^2} = 11$

$x - \dfrac{1}{x}$

Using identity

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

Substitute the value:

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 11 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 9 \\[1em] \Rightarrow x - \dfrac{1}{x} = \pm\sqrt{9} \\[1em] \Rightarrow x - \dfrac{1}{x} = \pm 3$

Since $x \gt 1, x$ is greater than $\dfrac{1}{x}$, so $x - \dfrac{1}{x}$ must be positive.

So, $x - \dfrac{1}{x} = 3$

$x^3 - \dfrac{1}{x^3}$

By using the identity:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Let $a = x$ and $b = \dfrac{1}{x}$.

$\Rightarrow x^3 - \dfrac{1}{x^3} = \Big(x - \dfrac{1}{x}\Big)\Big(x^2 + x \cdot \dfrac{1}{x} + \dfrac{1}{x^2}\Big) \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = \Big(x - \dfrac{1}{x}\Big)\Big(x^2 + \dfrac{1}{x^2} + 1\Big)\\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = (3)(11 + 1) \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = (3)(12) \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = 36$

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

If $\dfrac{1}{x + y} = \dfrac{1}{x} + \dfrac{1}{y}$ (x ≠ 0, y ≠ 0, x ≠ y), then find the value of x³ - y³

Answer

Given,

$\dfrac{1}{x + y} = \dfrac{1}{x} + \dfrac{1}{y}$

$\Rightarrow \dfrac{1}{x + y} = \dfrac{y + x}{xy} \\[1em] \Rightarrow xy = (x + y)(x + y) \\[1em] \Rightarrow xy = x^2 + y^2 + 2xy \\[1em] \Rightarrow 0 = x^2 + y^2 + 2xy - xy \\[1em] \Rightarrow x^2 + xy + y^2 = 0$

Using identity,

x³ - y³ = (x - y)(x² + xy +y²)

x³ - y³ = (x - y)(0)

x³ - y³ = 0

Hence, x³ - y³ = 0.

If $x^2 + y^2 + \dfrac{1}{x^2} + \dfrac{1}{y^2} = 4$, then find the value of x² + y²

Answer

Given,

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

Using identity,

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

$\Big(x - \dfrac{1}{x}\Big)^2 + \Big(y - \dfrac{1}{y}\Big)^2 = 0$

Hence,

$\Big(x - \dfrac{1}{x}\Big)^2 = 0 \text{ and } \Big(y - \dfrac{1}{y}\Big)^2 = 0$

$x - \dfrac{1}{x} = 0 \text{ and } y - \dfrac{1}{y} = 0$

Solving for x: $\Rightarrow x = \dfrac{1}{x} \\[1em] \Rightarrow x^2 = 1$

Solving for y: $\Rightarrow y = \dfrac{1}{y} \\[1em] \Rightarrow y^2 = 1$

Solving for x² + y²:

x² + y² = 1 + 1

x² + y² = 2

Hence, x² + y² = 2.