If x+$\frac{1}{x}$=$\sqrt{3}$ then find $x^3 + \frac{1}{x^3}$

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If x+$\frac{1}{x}$=$\sqrt{3}$ then find $x^3 + \frac{1}{x^3}$ [S.Chand Class ICSE 9 Chapter 3 Exercise 3.1 Expansion and Factorisation Solutions]

$x + \frac{1}{x} = \sqrt{3}$

$\therefore \left(x + \frac{1}{x}\right)^3 = \left(\sqrt{3}\right)^3$

$\Rightarrow x^3 + \frac{1}{x^3} + 3.x\ldotp \frac{1}{x}\left(x + \frac{1}{x}\right)$= $\sqrt{3} \times \sqrt{3} \times \sqrt{3}$

$\Rightarrow x^3 + \frac{1}{x^3} + 3\left(\sqrt{3}\right)$= $3\sqrt{3}$

$\Rightarrow x^3 + \frac{1}{x^3}$ = $3\sqrt{3} – 3\sqrt{3}$

$\Rightarrow x^3 + \frac{1}{x^3}$ = 0

$\therefore x^3 + \frac{1}{x^3}$ = 0

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