Simplify:

(i) (a + b)² + (a - b)²

(ii) (a + b)² - (a - b)²

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

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

(v) $\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{2b}{a} - \dfrac{a}{2b}\Big)^2$

(vi) $\Big(3x - \dfrac{1}{3x}\Big)^2 - \Big(3x + \dfrac{1}{3x}\Big)\Big(3x - \dfrac{1}{3x}\Big)$

(vii) (5a + 3b)² - (5a - 3b)² - 60ab

(viii) (3x + 1)² - (3x + 2)(3x - 1)

(i) Given,

⇒ (a + b)² + (a - b)²

⇒ a² + b² + 2ab + a² + b² - 2ab

⇒ 2a² + 2b²

⇒ 2(a² + b²)

Hence, (a + b)² + (a - b)² = 2(a² + b²).

(ii) Given,

⇒ (a + b)² - (a - b)²

⇒ (a² + b² + 2ab) - (a² + b² - 2ab)

⇒ a² + b² + 2ab - a² - b² + 2ab

⇒ 4ab

Hence, (a + b)² - (a - b)² = 4ab.

(iii) Given,

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

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

(iv) Given,

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

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

(v) Given,

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

Hence, $\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{2b}{a} - \dfrac{a}{2b}\Big)^2 = 4$.

(vi) Given,

$\Rightarrow \Big(3x - \dfrac{1}{3x}\Big)^2 - \Big(3x + \dfrac{1}{3x}\Big)\Big(3x - \dfrac{1}{3x}\Big) \\[1em] \Rightarrow \Big[(3x)^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times 3x \times \dfrac{1}{3x}\Big] - \Big[(3x)^2 - \Big(\dfrac{1}{3x}\Big)^2\Big] \\[1em] \Rightarrow \Big(9x^2 + \dfrac{1}{9x^2} - 2 \Big) - \Big(9x^2 - \dfrac{1}{9x^2}\Big) \\[1em] \Rightarrow 9x^2 + \dfrac{1}{9x^2} - 2 - 9x^2 + \dfrac{1}{9x^2} \\[1em] \Rightarrow \dfrac{2}{9x^2} - 2 \\[1em] \Rightarrow 2\Big(\dfrac{1}{9x^2} - 1\Big)$

Hence, $\Big(3x - \dfrac{1}{3x}\Big)^2 - \Big(3x + \dfrac{1}{3x})\Big(3x - \dfrac{1}{3x}\Big) = 2\Big(\dfrac{1}{9x^2} - 1\Big)$.

(vii) Given,

⇒ (5a + 3b)² - (5a - 3b)² - 60ab

⇒ [(5a)² + (3b)² + 2 × 5a × 3b] - [(5a)² + (3b)² - 2 × 5a × 3b] - 60ab

⇒ [25a² + 9b² + 2 × 5a × 3b] - [25a² + 9b² - 2 × 5a × 3b] - 60ab

⇒ 25a² + 9b² + 30ab - 25a² - 9b² + 30ab - 60ab

⇒ 60ab - 60ab

⇒ 0

Hence, (5a + 3b)² - (5a - 3b)² - 60ab = 0.

(viii) Given,

⇒ (3x + 1)² - [(3x + 2)(3x - 1)]

⇒ (3x)² + (1)² + 2 × 3x × 1 - (9x² - 3x + 6x - 2)

⇒ 9x² + 1 + 6x - (9x² + 3x - 2)

⇒ 9x² + 1 + 6x - 9x² - 3x + 2

⇒ 1 + 3x + 2

⇒ 3x + 3

⇒ 3(x + 1).

Hence, (3x + 1)² - (3x + 2)(3x - 1) = 3(x + 1).