# How do you simplify (x^3-1) / (x-1)?

Apr 14, 2016

$\frac{{x}^{3} - 1}{x - 1} = {x}^{2} + x - 1$ for $x \ne 1$

#### Explanation:

Note that $x = 1$ is not in the domain, as it would result in division by $0$. Also, using the difference of cubes formula , we have

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

Putting these together, we have, for $x \ne 1$,

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