# How do you solve  0= -n^3 + n?

Feb 14, 2017

$n = 0 , \text{ " n=-1," } n = 1$

#### Explanation:

Let's change the equation around a but first:

${n}^{3} - n = 0 \text{ } \leftarrow$ first term is positive.

$n \left({n}^{2} - 1\right) = 0 \text{ }$ factorise with a common factor

$n \left(n + 1\right) \left(n - 1\right) = 0 \text{ }$ difference of squares

Each factor can be equal to 0.

$n = 0 , \text{ " n+1=0" } n - 1 = 0$

Solutions are:

$n = 0 , \text{ " n=-1," } n = 1$