Answer
The solution set is $\left\{-2, 0, 2\right\}$.
Work Step by Step
Add $-8n$ to both sides to find:
$2n^3-8n=0$
Factor out $2n$ to find:
$2n(n^2-4)=0
\\2n(n^2-2^2)=0$
RECALL:
A difference of two squares can be factored using the formula:
$a^2-b^2=(a-b)(a+b)$
Factor the difference of two squares to find:
$2n(n-2)(n+2)=0$
Equate each factor to 0 then solve each equation to find:
$2n=0 \text{ or } n-2=0 \text{ or } n+2 = 0\\\
n=0 \text{ or } n=2 \text{ or } n=-2$
The solution set is $\left\{-2, 0, 2\right\}$.
