Expansions — Exercise 3.2

If x - y = 8 and xy = 5, find x² + y².

Answer

We know that,

(x - y)² = x² - 2xy + y²

⇒ x² + y² = (x - y)² + 2xy.

Substituting values we get,

⇒ x² + y² = (8)² + 2 × 5
⇒ x² + y² = 64 + 10
⇒ x² + y² = 74.

Hence, x² + y² = 74.

If x + y = 10 and xy = 21, find 2(x² + y²).

Answer

We know that,

(x + y)² = x² + 2xy + y²

⇒ x² + y² = (x + y)² - 2xy.

⇒ 2(x² + y²) = 2[(x + y)² - 2xy]

Substituting values we get,

⇒ 2(x² + y²) = 2[(10)² - 2 × 21]

⇒ 2(x² + y²) = 2(100 - 42)

⇒ 2(x² + y²) = 2 × 58

⇒ 2(x² + y²) = 116.

Hence, 2(x² + y²) = 116.

If 2a + 3b = 7 and ab = 2, find 4a² + 9b².

Answer

We know that,

a² + b² = (a + b)² - 2ab.

∴ 4a² + 9b² = (2a)² + (3b)² = (2a + 3b)² - 12ab.

Substituting values we get,

⇒ 4a² + 9b² = (7)² - 12 × 2

⇒ 4a² + 9b² = 49 - 24

⇒ 4a² + 9b² = 25.

Hence, 4a² + 9b² = 25.

If 3x - 4y = 16 and xy = 4, find the value of 9x² + 16y².

Answer

We know that,

a² + b² = (a - b)² + 2ab.

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

Substituting values we get,

⇒ 9x² + 16y² = (16)² + 24 × 4

⇒ 9x² + 16y² = 256 + 96

⇒ 9x² + 16y² = 352.

Hence, 9x² + 16y² = 352.

If x + y = 8 and x - y = 2, find the value of 2x² + 2y².

Answer

We know that,

(x + y)² = x² + y² + 2xy .....(i)

(x - y)² = x² + y² - 2xy ....(ii)

Adding eqn. (i) and (ii) we get,

(x + y)² + (x - y)² = x² + x² + y² + y² + 2xy - 2xy

= 2x² + 2y².

∴ 2x² + 2y² = (x + y)² + (x - y)².

Substituting values we get,

⇒ 2x² + 2y² = (8)² + (2)²

⇒ 2x² + 2y² = 64 + 4 = 68.

Hence, 2x² + 2y² = 68.

If a² + b² = 13 and ab = 6, find

(i) a + b

(ii) a - b

Answer

(i) We know that,

(a + b)² = a² + b² + 2ab

∴ (a + b) = $\sqrt{a^2 + b^2 + 2ab}$

Substituting values we get,

$\Rightarrow (a + b) = \sqrt{13 + 2 \times 6} \\[1em] \Rightarrow (a + b) = \sqrt{13 + 12} \\[1em] \Rightarrow (a + b) = \sqrt{25} \\[1em] \Rightarrow (a + b) = \pm 5.$

Hence, a + b = ±5.

(ii) We know that,

(a - b)² = a² + b² - 2ab

∴ (a - b) = $\sqrt{a^2 + b^2 - 2ab}$

Substituting values we get,

$\Rightarrow (a - b) = \sqrt{13 - 2 \times 6} \\[1em] \Rightarrow (a - b) = \sqrt{13 - 12} \\[1em] \Rightarrow (a - b) = \sqrt{1} \\[1em] \Rightarrow (a - b) = \pm 1.$

Hence, a - b = ±1.

If a + b = 4 and ab = -12, find

(i) a - b

(ii) a² - b².

Answer

(i) We know that,

(a - b)² = a² + b² - 2ab

(a - b)² = a² + b² + 2ab -2ab - 2ab

(a - b)² = (a + b)² - 4ab

(a - b) = $\sqrt{(a + b)^2 - 4ab}$

Substituting values we get,

$\Rightarrow (a - b) = \sqrt{(4)^2 - 4 \times (-12)} \\[1em] \Rightarrow (a - b) = \sqrt{16 + 48} \\[1em] \Rightarrow (a - b) = \sqrt{64} \\[1em] \Rightarrow (a - b) = \pm 8.$

Hence, a - b = ±8.

(ii) We know that,

a² - b² = (a + b)(a - b).

Substituting values we get,

⇒ a² - b² = 4 × ±8 = ±32.

Hence, a² - b² = ±32.

If p - q = 9 and pq = 36, evaluate

(i) p + q

(ii) p² - q².

Answer

(i) We know that,

(p - q)² = p² + q² - 2pq

⇒ (p - q)² = p² + q² + 2pq - 2pq - 2pq

⇒ (p - q)² = (p + q)² - 4pq

⇒ (p + q)² = (p - q)² + 4pq

⇒ (p + q) = $\sqrt{(p - q)^2 + 4pq}$

Substituting value we get,

$\Rightarrow p + q = \sqrt{9^2 + 4 \times 36} \\[1em] \Rightarrow p + q = \sqrt{81 + 144} \\[1em] \Rightarrow p + q = \sqrt{225} \\[1em] \Rightarrow p + q = \pm 15.$

Hence, p + q = ±15.

(ii) p² - q² = (p - q)(p + q).

Substituting value we get,

⇒ p² - q² = 9 × ±15 = ±135.

Hence, p² - q² = ±135.

If x + y = 6 and x - y = 4, find

(i) x² + y²

(ii) xy.

Answer

(i) We know that,

(x + y)² = x² + y² + 2xy .....(i)

(x - y)² = x² + y² - 2xy ....(ii)

Adding eqn. (i) and (ii) we get,

(x + y)² + (x - y)² = x² + x² + y² + y² + 2xy - 2xy = 2x² + 2y².

⇒ 2x² + 2y² = (x + y)² + (x - y)².

∴ x² + y² = $\dfrac{(x + y)^2 + (x - y)^2}{2}$

Substituting values we get,

$\Rightarrow x^2 + y^2 = \dfrac{(6)^2 + (4)^2}{2} \\[1em] = \dfrac{36 + 16}{2} \\[1em] = \dfrac{52}{2} \\[1em] = 26.$

Hence, x² + y² = 26.

(ii) We know that,

(x + y)² = x² + y² + 2xy .....(i)

(x - y)² = x² + y² - 2xy ....(ii)

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

(x + y)² - (x - y)² = x² - x² + y² - y² + 2xy - (-2xy) = 4xy.

⇒ (x + y)² - (x - y)² = 4xy.

∴ xy = $\dfrac{(x + y)^2 - (x - y)^2}{4}$

Substituting values we get,

$xy = \dfrac{6^2 - 4^2}{4} \\[1em] = \dfrac{36 - 16}{4} \\[1em] = \dfrac{20}{4} \\[1em] = 5.$

Hence, xy = 5.

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

Answer

Given,

$\phantom{\therefore} x - 3 = \dfrac{1}{x} \\[1em] \therefore x - \dfrac{1}{x} = 3 \\[1em]$

We know that,

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

Substituting values we get,

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

Hence, $x^2 + \dfrac{1}{x^2}$ = 7.

If x + y = 8 and xy = $3\dfrac{3}{4}$, find the values of

(i) x - y

(ii) 3(x² + y²)

(iii) 5(x² + y²) + 4(x - y).

Answer

(i) We know that,

(x + y)² = x² + y² + 2xy .....(i)

(x - y)² = x² + y² - 2xy .....(ii)

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

(x + y)² - (x - y)² = x² - x² + y² - y² + 2xy - (-2xy) = 4xy.

⇒ (x + y)² - (x - y)² = 4xy.

∴ (x - y) = $\sqrt{(x + y)^2 - 4xy}$.

Substituting values we get,

$x - y = \sqrt{8^2 - 4 \times 3\dfrac{3}{4}} \\[1em] = \sqrt{64 - 4 \times \dfrac{15}{4}} \\[1em] = \sqrt{64 - 15} \\[1em] = \sqrt{49} \\[1em] = \pm 7.$

Hence, x - y = ±7.

(ii) We know that,

3(x² + y²) = 3[(x + y)² - 2xy].

Substituting values we get,

$3(x^2 + y^2) = 3\Big[(8)^2 - 2 \times 3\dfrac{3}{4}\Big] \\[1em] = 3\Big(64 - 2 \times \dfrac{15}{4}\Big) \\[1em] = 3\Big(64 - \dfrac{15}{2}\Big) \\[1em] = 3\Big(\dfrac{128 - 15}{2}\Big) \\[1em] = 3 \times \dfrac{113}{2} \\[1em] = \dfrac{339}{2} \\[1em] = 169\dfrac{1}{2}.$

Hence, 3(x² + y²) = $169\dfrac{1}{2}.$

(iii) From parts (i) and (ii) we get,

(x - y) = ±7 and x² + y² = $\dfrac{113}{2}$.

When (x - y) = 7,

Substituting values we get,

$5(x^2 + y^2) + 4(x - y) = 5 \times \dfrac{113}{2} + 4 \times 7 \\[1em] = \dfrac{565}{2} + 28 \\[1em] = \dfrac{565 + 56}{2} \\[1em] = \dfrac{621}{2} \\[1em] = 310\dfrac{1}{2}.$

When (x - y) = -7,

Substituting values we get,

$5(x^2 + y^2) + 4(x - y) = 5 \times \dfrac{113}{2} + 4 \times -7 \\[1em] = \dfrac{565}{2} - 28 \\[1em] = \dfrac{565 - 56}{2} \\[1em] = \dfrac{509}{2} \\[1em] = 254\dfrac{1}{2}.$

Hence, 5(x² + y²) + 4(x - y) = $310\dfrac{1}{2} \text{ or } 254\dfrac{1}{2}$.

If x² + y² = 34 and xy = $10\dfrac{1}{2}$, find the value of 2(x + y)² + (x - y)².

Answer

We know that,

$2(x + y)^2 + (x - y)^2 = 2(x^2 + y^2 + 2xy) + (x^2 + y^2 - 2xy) \\[1em] = 2x^2 + 2y^2 + 4xy + x^2 + y^2 - 2xy \\[1em] = 3x^2 + 3y^2 + 2xy \\[1em] = 3(x^2 + y^2) + 2xy.$

Substituting values in above equation,

$= 3 \times 34 + 2 \times 10\dfrac{1}{2} \\[1em] = 102 + 2 \times \dfrac{21}{2} \\[1em] = 102 + 21 \\[1em] = 123.$

Hence, 2(x + y)² + (x - y)² = 123.

If a - b = 3 and ab = 4, find a³ - b³.

Answer

We know that,

⇒ (a - b)³ = a³ - b³ - 3ab(a - b).

∴ a³ - b³ = (a - b)³ + 3ab(a - b).

Substituting values we get,

⇒ a³ - b³ = (3)³ + 3 × 4 × 3
⇒ a³ - b³ = 27 + 36 = 63.

Hence, a³ - b³ = 63.

If 2a - 3b = 3 and ab = 2, find the value of 8a³ - 27b³.

Answer

We know that,

⇒ (a - b)³ = a³ - b³ - 3ab(a - b).

∴ a³ - b³ = (a - b)³ + 3ab(a - b).

∴ 8a³ - 27b³ = (2a)³ - (3b)³ = (2a - 3b)³ + 3 × 2a × 3b (2a - 3b)

Substituting values we get,

⇒ 8a³ - 27b³ = (2a - 3b)³ + 3 × 2a × 3b (2a - 3b)

⇒ 8a³ - 27b³ = 3³ + 18ab × 3

⇒ 8a³ - 27b³ = 27 + 18 × 2 × 3

⇒ 8a³ - 27b³ = 27 + 108

⇒ 8a³ - 27b³ = 135.

Hence, 8a³ - 27b³ = 135.

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

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

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

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

(iv) $x - \dfrac{1}{x}$.

Answer

(i) We know that,

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

Substituting values we get,

$x^2 + \dfrac{1}{x^2} = 4^2 - 2 \\[1em] = 16 - 2 \\[1em] = 14.$

Hence, $x^2 + \dfrac{1}{x^2}$ = 14.

(ii) We know that,

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

Substituting values we get,

$x^4 + \dfrac{1}{x^4} = 14^2 - 2 \\[1em] = 196 - 2 \\[1em] = 194.$

Hence, $x^4 + \dfrac{1}{x^4}$ = 194.

(iii) We know that,

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

Substituting values we get,

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

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

(iv) We know that,

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

Substituting values we get,

$x - \dfrac{1}{x} = \sqrt{14 - 2} \\[1em] = \sqrt{12} \\[1em] = \sqrt{4 \times 3} \\[1em] = \pm 2\sqrt{3}.$

Hence, $x - \dfrac{1}{x} = \pm 2\sqrt{3}$.

If $x - \dfrac{1}{x} = 5$, find the value of $x^4 + \dfrac{1}{x^4}$.

Answer

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

Putting this value of $(x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2$ in eqn (i), we get:

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

Putting this value of $x^2 + \dfrac{1}{x^2}$ in eqn (ii), we get:

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

Putting the given values in above eqn, we get:

$x^4 + \dfrac{1}{x^4} = (5^2 + 2)^2 - 2 \\[1em] = (27)^2 - 2 \\[1em] = 729 - 2 \\[1em] = 727.$

Hence, $x^4 + \dfrac{1}{x^4}$ = 727.

If $x - \dfrac{1}{x} = \sqrt{5}$, find the values of

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

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

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

Answer

(i) We know that,

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

Substituting values we get,

$x^2 + \dfrac{1}{x^2} = (\sqrt{5})^2 + 2 \\[1em] = 5 + 2 \\[1em] = 7.$

Hence, $x^2 + \dfrac{1}{x^2}$ = 7.

(ii) We know that,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \therefore x + \dfrac{1}{x} = \sqrt{x^2 + \dfrac{1}{x^2} + 2}$

Substituting values we get,

$x + \dfrac{1}{x} = \sqrt{7 + 2} \\[1em] = \sqrt{9} \\[1em] = \pm 3.$

Hence, $x + \dfrac{1}{x} = \pm 3.$

(iii) We know that,

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

When $x + \dfrac{1}{x} = 3$.

Substituting values we get,

$x^3 + \dfrac{1}{x^3} = 3^3 - 3 \times 3 \\[1em] = 27 - 9 \\[1em] = 18.$

When $x + \dfrac{1}{x} = -3$.

Substituting values we get,

$x^3 + \dfrac{1}{x^3} = (-3)^3 - 3 \times (-3) \\[1em] = -27 + 9 \\[1em] = -18.$

Hence, $x^3 + \dfrac{1}{x^3} = \pm 18$.

If $x + \dfrac{1}{x}$ = 6, find

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

(ii) $x^2 - \dfrac{1}{x^2}$

Answer

(i) We know that,

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

Eq. (i) can be written as,

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

Substituting values we get,

$x - \dfrac{1}{x} = \sqrt{(6)^2 - 4} \\[1em] = \sqrt{36 - 4} \\[1em] = \sqrt{32} \\[1em] = \pm 4\sqrt{2}.$

Hence, $x - \dfrac{1}{x} = \pm 4\sqrt{2}.$

(ii) We know that,

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

When, $x - \dfrac{1}{x} = 4\sqrt{2}$

Substituting values we get,

$x^2 - \dfrac{1}{x^2} = 6 \times 4\sqrt{2} = 24\sqrt{2}.$

When, $x - \dfrac{1}{x} = -4\sqrt{2}$

Substituting values we get,

$x^2 - \dfrac{1}{x^2} = 6 \times -4\sqrt{2} = -24\sqrt{2}.$

Hence, $x^2 - \dfrac{1}{x^2} = ±24\sqrt{2}.$

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

Answer

We know that,

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

Substituting values we get,

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

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,

$x^3 + \dfrac{1}{x^3} = (2)^3 - 3 \times 2 = 8 - 6 = 2$ .......(ii)

We know that,

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

Substituting values we get,

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

From (i), (ii) and (iii),

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

If $x - \dfrac{2}{x} = 3$, find the value of $x^3 - \dfrac{8}{x^3}$.

Answer

We know that,

⇒ (a - b)³ = a³ - b³ - 3ab(a - b)

⇒ a³ - b³ = (a - b)³ + 3ab(a - b).

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

Substituting values we get,

$x^3 - \dfrac{8}{x^3} = 3^3 + 6 \times 3 = 27 + 18 = 45.$

Hence, the value of $x^3 - \dfrac{8}{x^3} = 45.$

If a + 2b = 5, prove that a³ + 8b³ + 30ab = 125.

Answer

We know that,

⇒ (a + 2b)³ = a³ + 8b³ + 3(a)(2b)(a + 2b)

Substituting values in above formula,

⇒ 5³ = a³ + 8b³ + 6ab × 5
⇒ 125 = a³ + 8b³ + 30ab.

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

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

Answer

We know that,

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

Substituting values we get,

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

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

If $x^2 + \dfrac{1}{x^2} = 27$, find the value of $3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x}.$

Answer

We know that,

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

Substituting values we get,

$x - \dfrac{1}{x} = \sqrt{27 - 2} = \sqrt{25} = \pm 5.$

Here, $3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x}$ can be written as,

$3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = 3\Big(x^3 - \dfrac{1}{x^3}\Big) + 5\Big(x - \dfrac{1}{x}\Big) \\[1em] = 3\Big[\Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big)\Big] + 5\Big(x - \dfrac{1}{x}\Big).$

Considering $x - \dfrac{1}{x} = 5$ in first case and substituting value we get,

$3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = 3(5^3 + 3 \times 5) + (5 \times 5) \\[1em] = 3(125 + 15) + 25 \\[1em] = (3 \times 140) + 25 \\[1em] = 420 + 25 \\[1em] = 445.$

Considering $x - \dfrac{1}{x} = -5$ in second case and substituting value we get,

$3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = 3[(-5)^3 + 3 \times (-5)] + [5 \times (-5)] \\[1em] = 3(-125 - 15) - 25 \\[1em] = (3 \times -140) - 25 \\[1em] = -420 - 25 \\[1em] = -445.$

Hence, $3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = \pm 445.$

If $x^2 + \dfrac{1}{25x^2} = 8\dfrac{3}{5}, \text{ find } x + \dfrac{1}{5x}$.

Answer

$x^2 + \dfrac{1}{25x^2} = (x)^2 + \dfrac{1}{(5x)^2} ....[i] \\[1em] \Big(x + \dfrac{1}{5x}\Big)^2 = (x)^2 + 2 \times x \times \dfrac{1}{5x} + \dfrac{1}{(5x)^2} \\[1em] = (x)^2 + \dfrac{1}{(5x)^2} + \dfrac{2}{5} \\[1em] \therefore x^2 + \dfrac{1}{25x^2} = \Big(x + \dfrac{1}{5x}\Big)^2 - \dfrac{2}{5}$

Putting this value of $x^2 + \dfrac{1}{25x^2}$ in eqn (i), we get:

$x^2 + \dfrac{1}{25x^2} = \Big(x + \dfrac{1}{5x}\Big)^2 - \dfrac{2}{5}$

Let $x + \dfrac{1}{5x}$ be a.

Substituting values we get,

$\Rightarrow 8\dfrac{3}{5} = a^2 - \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{43}{5} = \dfrac{5a^2 - 2}{5} \\[1em] \Rightarrow 5a^2 - 2 = 43 \\[1em] \Rightarrow 5a^2 = 45 \\[1em] \Rightarrow a^2 = 9 \\[1em] \Rightarrow a = \pm 3.$

Hence, $x + \dfrac{1}{5x} = \pm 3$.

If $x^2 + \dfrac{1}{4x^2} = 8, \text{ find } x^3 + \dfrac{1}{8x^3}$.

Answer

We know that,

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

Substituting values in above equation we get,

$\Rightarrow 8 = \Big(x + \dfrac{1}{2x}\Big)^2 - 1 \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = 8 + 1 \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = 9 \\[1em] \Rightarrow x + \dfrac{1}{2x} = \sqrt{9} \\[1em] \Rightarrow x + \dfrac{1}{2x} = \pm 3.$

We know that,

$\Rightarrow (a + b)^3 = a^3 + b^3 + 3ab(a + b) \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^3 = x^3 + \Big(\dfrac{1}{2x}\Big)^3 + 3 \times x \times \dfrac{1}{2x}\Big(x + \dfrac{1}{2x}\Big) \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^3 = x^3 + \dfrac{1}{8x^3} + \dfrac{3}{2}\Big(x + \dfrac{1}{2x}\Big) \\[1em] \therefore x^3 + \dfrac{1}{8x^3} = \Big(x + \dfrac{1}{2x}\Big)^3 - \dfrac{3}{2}\Big(x + \dfrac{1}{2x}\Big).$.

Case 1: $x + \dfrac{1}{2x} = 3$ substituting values we get,

$x^3 + \dfrac{1}{8x^3} = 3^3 - \dfrac{3}{2} \times 3 \\[1em] = 27 - \dfrac{9}{2} \\[1em] = \dfrac{54 - 9}{2} \\[1em] = \dfrac{45}{2}.$

Case 2 : $x + \dfrac{1}{2x} = -3$, substituting values we get,

$x^3 + \dfrac{1}{8x^3} = (-3)^3 - \dfrac{3}{2} \times (-3) \\[1em] = -27 + \dfrac{9}{2} \\[1em] = \dfrac{-54 + 9}{2} \\[1em] = -\dfrac{45}{2}.$

Hence, $x^3 + \dfrac{1}{8x^3} = \pm \dfrac{45}{2} = \pm 22\dfrac{1}{2}$.

If a² - 3a + 1 = 0, find

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

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

Answer

Dividing each term of a² - 3a + 1 = 0 by a we get,

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

(i) We know that,

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

Substituting values we get,

$a^2 + \dfrac{1}{a^2} = 3^2 - 2 = 9 - 2 = 7.$

Hence, $a^2 + \dfrac{1}{a^2}$ = 7.

(ii) We know that,

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

Substituting values we get,

$a^3 + \dfrac{1}{a^3} = 3^3 - 3 \times 3 = 27 - 9 = 18.$

Hence, $a^3 + \dfrac{1}{a^3} = 18.$

If a = $\dfrac{1}{a - 5}$, find

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

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

(iii) $a^2 - \dfrac{1}{a^2}$

Answer

Given,

$\Rightarrow a = \dfrac{1}{a - 5} \\[1em] \therefore a(a - 5) = 1 \\[1em] \Rightarrow a^2 - 5a = 1 \\[1em] \Rightarrow a^2 - 5a - 1 = 0$

Now,

(i) Dividing above equation by a we get,

$\Rightarrow a - 5 - \dfrac{1}{a} = 0 \\[1em] \Rightarrow a - \dfrac{1}{a} = 5.$.

Hence, $a - \dfrac{1}{a} = 5$.

(ii) We know that,

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

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

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

Substituting values we get,

$a + \dfrac{1}{a} = \sqrt{5^2 + 4} \\[1em] = \sqrt{25 + 4} \\[1em] = \pm \sqrt{29}.$

Hence, $a + \dfrac{1}{a} = \pm \sqrt{29}$.

(iii) We know that,

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

Substituting values we get,

$\Big(a^2 - \dfrac{1}{a^2}\Big) = \pm \sqrt{29} \times 5 = \pm 5\sqrt{29}.$

Hence, $a^2 - \dfrac{1}{a^2} = \pm 5\sqrt{29}$.

If $\Big(x + \dfrac{1}{x}\Big)^2 = 3, \text{ find } x^3 + \dfrac{1}{x^3}.$

Answer

Given,

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

We know that,

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

When $x + \dfrac{1}{x} = \sqrt 3$ substituting values we get,

$x^3 + \dfrac{1}{x^3} = (\sqrt{3})^3 - 3 \times \sqrt{3} = 3\sqrt{3} - 3\sqrt{3} = 0.$

When $x + \dfrac{1}{x} = -\sqrt{3}$ substituting values we get,

$x^3 + \dfrac{1}{x^3} = (-\sqrt{3})^3 - 3 \times (-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0.$

Hence, $x^3 + \dfrac{1}{x^3} = 0$.

If $x = 5 - 2\sqrt{6}$, find the value of $\sqrt{x} + \dfrac{1}{\sqrt{x}}$

Answer

Given,

$x = 5 - 2\sqrt{6} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{5- 2\sqrt{6}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}} \times \dfrac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{(5 - 2\sqrt{6})(5 + 2\sqrt{6})} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{25 - 24} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{1} = 5 + 2\sqrt{6}.$

So,

$x + \dfrac{1}{x} = 5 - 2\sqrt{6} + 5 + 2\sqrt{6} = 10.$

We know that,

$\Rightarrow \Big(\sqrt{x} + \dfrac{1}{\sqrt{x}}\Big)^2 = x + \dfrac{1}{x} + 2 \times \sqrt{x} \times \dfrac{1}{\sqrt{x}} \\[1em] \Rightarrow \Big(\sqrt{x} + \dfrac{1}{\sqrt{x}}\Big)^2 = x + \dfrac{1}{x} + 2 \\[1em] \Rightarrow \sqrt{x} + \dfrac{1}{\sqrt{x}} = \sqrt{x + \dfrac{1}{x} + 2} = \sqrt{10 + 2} = \sqrt{12} = \pm 2\sqrt{3}.$

Hence, $\sqrt{x} + \dfrac{1}{\sqrt{x}} = \pm 2\sqrt{3}.$

If a + b + c = 12 and ab + bc + ca = 22, find a² + b² + c².

Answer

We know that,

a² + b² + c² = (a + b + c)² - 2(ab + bc + ca)

Substituting values we get,

a² + b² + c² = (12)² - 2(22) = 144 - 44 = 100.

Hence, a² + b² + c² = 100.

If a + b + c = 12 and a² + b² + c² = 100, find ab + bc + ca.

Answer

We know that,

a² + b² + c² = (a + b + c)² - 2(ab + bc + ca)

Substituting values we get,

⇒ 100 = 12² - 2(ab + bc + ca)

⇒ 100 = 144 - 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 144 - 100

⇒ 2(ab + bc + ca) = 44

⇒ (ab + bc + ca) = 22.

Hence, ab + bc + ca = 22.

If a² + b² + c² = 125 and ab + bc + ca = 50, find a + b + c.

Answer

We know that,

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

Substituting values we get,

⇒ (a + b + c)² = 125 + 2(50) = 125 + 100 = 225.

$\therefore (a + b + c) = \sqrt{225} = \pm 15.$

Hence, (a + b + c) = ± 15.

If a + b - c = 5 and a² + b² + c² = 29, find the value of ab - bc - ca.

Answer

Given, a + b - c = 5. Squaring both sides we get,

⇒ (a + b - c)² = 5²

⇒ a² + b² + c² + 2ab - 2bc - 2ca = 25

Substituting values we get,

⇒ 29 + 2(ab - bc - ca) = 25

⇒ 2(ab - bc - ca) = 25 - 29

⇒ 2(ab - bc - ca) = -4

⇒ (ab - bc - ca) = -2.

Hence, (ab - bc - ca) = -2.

If a - b = 7 and a² + b² = 85, then find the value of a³ - b³.

Answer

We know that,

⇒ (a - b)² = a² + b² - 2ab

⇒ (7)² = 85 - 2ab

⇒ 49 = 85 - 2ab

⇒ 2ab = 85 - 49

⇒ 2ab = 36

⇒ ab = 18.

We know that,

a³ - b³ = (a - b)(a² + b² + ab)

Substituting values we get,

⇒ a³ - b³ = 7(85 + 18) = 7(103) = 721.

Hence, a³ - b³ = 721.

If the number x is 3 less than the number y and the sum of the squares of x and y is 29, find the product of x and y.

Answer

Given,

x = y - 3 or x - y = -3 and

x² + y² = 29.

We know that,

⇒ (x + y)² = x² + y² + 2xy ......(i)

⇒ (x - y)² = x² + y² - 2xy .......(ii)

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

⇒ (x + y)² - (x - y)² = x² + y² + 2xy - (x² + y² - 2xy)

⇒ (x + y)² - (x - y)² = 4xy

⇒ x² + y² = (x - y)² + 4xy

Substituting values we get,

⇒ 29 = (-3)² + 4xy
⇒ 29 = 9 + 4xy
⇒ 29 - 9 = 4xy
⇒ 4xy = 20
⇒ xy = 5.

Hence, xy = 5.

If the sum and the product of two numbers are 8 and 15 respectively, find the sum of their cubes.

Answer

Let two numbers be x and y.

Given,

x + y = 8 and xy = 15.

We know that,

⇒ x³ + y³ = (x + y)³ - 3xy(x + y)

⇒ x³ + y³ = 8³ - 3(15)(8)

⇒ x³ + y³ = 512 - 360

⇒ x³ + y³ = 152.

Hence, x³ + y³ = 152.