Question

Simplify:
$(x + \frac{1}{x} ) {}^{2} - (x - \frac{1}{x} ) {}^{2}$
​

Answer 1

Solution :-

→ (x + 1/x)² - (x - 1/x)²

using :-

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab

→ (x² + 1/x² + 2 * x * 1/x) - (x² + 1/x² - 2 * x * 1/x)

→ (x² + 1/x² + 2) - (x² + 1/x² - 2)

→ x² - x² + 1/x² - 1/x² + 2 + 2

→ 4 (Ans.)

_________________

Shortcut :-

• (a + b)² - (a - b)² = 4ab

we have :-

• a = x
• b = (1/x)

So,

→ (x + 1/x)² - (x - 1/x)²

→ 4 * x * (1/x)

→ 4 (Ans.)

__________________________

Answer 2

$\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}}$

• We can solve this by 2 methods

$\sf{\red{\boxed{\bold{Long\:method}}}}$

$\sf\implies ( x + \dfrac{1}{x})^2 - ( x - \dfrac{1}{x})^2$

$\sf{\bold{\green{\underline{\underline{( a + b)^2 = a^2 + b^2 + 2ab }}}}}$

$\sf{\bold{\green{\underline{\underline{( a-b)^2 = a^2 + b^2 - 2ab }}}}}$

$\sf\implies ( x^2 + (\dfrac{1}{x})^2 + 2 ) - ( x^2 - (\dfrac{1}{x})^2 - 2 )$

$\sf\implies x^2 + \dfrac{1}{x}^2 + 2 - x - \dfrac{1}{x}^2 + 2$

$\sf\implies \cancel{x^2} +\cancel{ \dfrac{1}{x}^2} + 2 - \cancel{x} - \cancel{\dfrac{1}{x}^2} + 2$

$\sf\implies 2 + 2$

$\sf\implies 4$

$\longrightarrow\sf{\bold{\orange{\underline{\underline{\dag 4 }}}}}$

$\sf{\red{\boxed{\bold{Shortcut\: method}}}}$

$\sf\implies ( x + \dfrac{1}{x})^2 - ( x - \dfrac{1}{x})^2$

$\sf{\bold{\green{\underline{\underline{( a + b )^2 - ( a - b)^2 = 4ab}}}}}$

$\sf\implies 4 \times x \times\dfrac{1}{x}$

$\sf\implies 4$

$\longrightarrow\sf{\bold{\orange{\underline{\underline{\dag 4}}}}}$