Question

People please solve this ASAP
$x - \frac{1}{x} = - \sqrt{3} \\ \\ x^{3} - \frac{1}{x^{3} } =$
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Answer 1

$\mathrm{Given :\;x - \dfrac{1}{x} = -\sqrt{3}}$

Cubing on both sides

$\mathrm{\implies \left(x - \dfrac{1}{x}\right)^3 = (-\sqrt{3})^3}$

Use Identity : (p - q)³ = p³ - 3p²q + 3pq² + q³

$\mathrm{\implies (x)^3 - \left(\dfrac{1}{x}\right)^3 + 3(x)\left(\dfrac{1}{x}\right)^2 - 3(x)^2\left(\dfrac{1}{x}\right)= -3\sqrt{3}}$

$\mathrm{\implies x^3 - \dfrac{1}{x^3} + 3x\left(\dfrac{1}{x^2}\right) - 3x^2\left(\dfrac{1}{x}\right)= -3\sqrt{3}}$

$\mathrm{\implies x^3 - \dfrac{1}{x^3} + \dfrac{3}{x} - 3x = -3\sqrt{3}}$

$\mathrm{\implies x^3 - \dfrac{1}{x^3} - 3\left(x - \dfrac{1}{x}\right) = -3\sqrt{3}}$

$\mathrm{\implies x^3 - \dfrac{1}{x^3} - 3\left(-\sqrt{3}\right) = -3\sqrt{3}}$

$\mathrm{\implies x^3 - \dfrac{1}{x^3} + 3\sqrt{3} = -3\sqrt{3}}$

$\mathrm{\implies x^3 - \dfrac{1}{x^3} = -3\sqrt{3} - 3\sqrt{3}}$

$\mathrm{\implies x^3 - \dfrac{1}{x^3} = -6\sqrt{3}}$