Answer
The solution set is $\left\{-\frac{1}{2}, 0, \frac{1}{2}\right\}$.
Work Step by Step
Add $-9x$ to both sides to find:
$36x^3-9x=0$
Factor out $9x$ to find:
$9x(4x^2-1)=0
\\=9x[(2x)^2-1^{2}]=0$
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:
$9x(2x-1)(2x+1)=0$
Equate each factor to 0 then solve each equation to find:
$9x=0 \text{ or } 2x-1=0 \text{ or } 2x+1=0
\\x=0 \text{ or } 2x=1 \text{ or } 2x=-1
\\x=0 \text{ or } x=\frac{1}{2} \text{ or } x=-\frac{1}{2}$
The solution set is $\left\{-\frac{1}{2}, 0, \frac{1}{2}\right\}$.
