Answer
The solution set is $\left\{-\frac{1}{4}, 0, \frac{1}{4}\right\}$.
Work Step by Step
Add $-4x$ to both sides to find:
$64x^3-4x=0$
Factor out $4x$ to find:
$4x(16x^2-1)=0
\\=4x[(4x)^2-1^{2}]=0$
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:
$4x(4x-1)(4x+1)=0$
Equate each factor to 0 then solve each equation to find:
$4x=0 \text{ or } 4x-1=0 \text{ or } 4x+1=0
\\x=0 \text{ or } 4x=1 \text{ or } 4x=-1
\\x=0 \text{ or } x=\frac{1}{4} \text{ or } x=-\frac{1}{4}$
The solution set is $\left\{-\frac{1}{4}, 0, \frac{1}{4}\right\}$.
