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Algebraic Identities

Question

If ${(x + \frac{1}{x})}^{2} - 2 = 7$ . Find the value of $x^{3} + \frac{1}{x^{3}}$ .

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Solution

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$\Rightarrow {(x + \frac{1}{x})}^{2} - 2 = 7$

$\Rightarrow {(x + \frac{1}{x})}^{2} = 9$

$\Rightarrow (x + \frac{1}{x}) = 3$

Cube on both sides

$\Rightarrow x^{3} + \frac{1}{x^{3}} + 3 x \frac{1}{x} (x + \frac{1}{x}) = 27$

$\Rightarrow x^{3} + \frac{1}{x^{3}} + 3 (3) = 27$

$\Rightarrow x^{3} + \frac{1}{x^{3}} = 18$

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