Question

Please solve this question​

Answer 1

$\bf{\large{\underline{\underline{Question:-}}}}$

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

$\bf{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\bf{{x}^{3} - \dfrac{1}{ {x}^{3}} = 140}}$

$\bf{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\sf{x = \dfrac{1}{x - 5}}$

To find :- $\sf{x^3 + \dfrac{1}{x^3}}$

Solution :-

$\tt{x = \dfrac{1}{x - 5}}$

$\tt{\implies{ \dfrac{1}{x} = x - 5}}$

$\tt{\implies{0 = x - 5 - \dfrac{1}{x} }}$

$\tt{\implies{5 = x - \dfrac{1}{x} }}$

$\tt{\implies{x - \dfrac{1}{x} = 5}}$

By cubing on both the sides :-

$\tt{\implies{{(x - \dfrac{1}{x})}^3 = 5^3}}$

We know that (x - y)³ = x³ - y³ - 3xy(x - y)

Here x = x, y = 1/x

By substituting the values in the identity we have,

$\tt{\implies{{x}^{3} - \dfrac{1}{ {x}^{3}} - 3(x)( \dfrac{1}{x} )(x - \dfrac{1}{x}) = 125}}$

$\tt{\implies{{x}^{3} - \dfrac{1}{ {x}^{3}} - 3(x - \dfrac{1}{x}) = 125}}$

$\tt{\implies{{x}^{3} - \dfrac{1}{ {x}^{3}} - 3(5) = 125}}$

[Since x - 1/x = 5]

$\tt{\implies{{x}^{3} - \dfrac{1}{ {x}^{3}} - 15 = 125}}$

$\tt{\implies{{x}^{3} - \dfrac{1}{ {x}^{3}} = 125 + 15}}$

$\tt{\implies{{x}^{3} - \dfrac{1}{ {x}^{3}} = 140}}$

$\boxed{\bf{{x}^{3} - \dfrac{1}{ {x}^{3}} = 140}}$

$\bf{\large{\underline{\underline{Identity\:Used:-}}}}$

(x - y)³ = x³ - y³ - 3xy(x - y)

$\bf{\large{\underline{\underline{Some\:Important\:Identities:-}}}}$

[1] (x + y)² = x² + y² + 2xy

[2] (x - y)² = x² + y² - 2xy

[3] (x + y)(x - y) = x² - y²

[4] (x + a)(x + b) = x² + (a + b)x + ab