Answer
The solution set is $\left\{-4, 0, 4\right\}$.
Work Step by Step
Factor out $3x$ to obtain:
$3x(x^2-16)=0
\\3x(x^2-4^2)=0
\\3x(x-4)(x+4)=0$
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 obtain:
$3x(x-4)(x+4)=0$
Equate each factor to 0, and then solve each equation to obtain:
$3x=0 \text{ or } x-4=0 \text{ or } x+4 = 0\\\
x=0 \text{ or } x=4 \text{ or } x=-4$
The solution set is $\left\{-4, 0, 4\right\}$.
