Question

Factorize the expression $x^3-\frac{1}{x^3}-2x+\frac{2}{x}$.

Answer 1

Factorize the given expression.

The given expression is $x^3-\frac{1}{x^3}-2x+\frac{2}{x}$.

It can be rewritten as:

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

Now use the identity, $a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$ to expand.

$$\begin{gathered} {\left(x\right)}^3-{\left(\frac{1}{x}\right)}^3-2\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)\left[x^2+\left(x\right)\left(\frac{1}{x}\right)+{\left(\frac{1}{x}\right)}^2\right]-2\left(x-\frac{1}{x}\right) \\ =\left(x-\frac{1}{x}\right)\left(x^2+1+\frac{1}{x^2}\right)-2\left(x-\frac{1}{x}\right) \\ =\left(x-\frac{1}{x}\right)\left(x^2+1+\frac{1}{x^2}-2\right) \\ =\left(x-\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}\right) \end{gathered}$$

Hence, the expression, $x^3-\frac{1}{x^3}-2x+\frac{2}{x}=\left(x-\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}\right)$.