An algebra problem by A Former Brilliant Member

$\begin{aligned} \left(x - \frac 1x \right)^3 & = x^3 - 3x + \frac 3x - \frac 1{x^3} \\ 1^3 & = x^3 - \frac 1{x^3} - 3 \left(x - \frac 1x \right) \\ 1^3 & = x^3 - \frac 1{x^3} - 3(1) \\ \implies x^3 - \frac 1{x^3} & = \boxed{4} \end{aligned}$

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

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

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

$x^3-\dfrac{3x^2}{x}+\dfrac{3x}{x^2}-\dfrac{1}{x^3}=1$

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

$x^3-3\left(x-\dfrac{1}{x}\right)-\dfrac{1}{x^3}=1$ but: $x-\dfrac{1}{x}=1$

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

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

$x^3-\dfrac{1}{x^3}=4$

Given that,

• $x -$ $\dfrac{1}{x}$ $= 1$

Now,

$x^3$ $-$ $\dfrac{1}{x^3}$

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

$= (x -$ $\dfrac{1}{x})^3$ $+$ $3 \cdot$ $x$ $\dfrac{1}{x}$ $(x -$ $\dfrac{1}{x})$

$= (1)^3 + 3 \times 1$ ;[ $x$ $\dfrac{1}{x}$ cancelled]

Therefore , $x^3$ - $\dfrac{1}{x^3}$ $= 4$

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

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

$\implies x^3 - \dfrac{1}{x^3} = \boxed{4}$ .