Answer
The solution set is $\left\{-3, 0, 3\right\}$.
Work Step by Step
Factor out $5c$ to obtain:
\begin{align*}
5c(c^2-9)&=0\\
5c(c-3)(c+3)&=0
\end{align*}
Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain:
\begin{align*}
5c&=0 &\text{or}& &c-3=0& &\text{or}& &c+3=0\\
c&=0 &\text{or}& &c=3& &\text{or}& &c=-3\\
\end{align*}
Checking:
\begin{align*}
5(0^3)-45(0)&=0\\
5(0)-0&=0\\
0&\stackrel{\checkmark}=0
\end{align*}
\begin{align*}
5(3^3)-45(3)&=0\\
5(27)-135&=0\\
135-135&=0\\
0&\stackrel{\checkmark}=0
\end{align*}
\begin{align*}
5(-3)^3-45(-3)&=0\\
5(-27)+135&=0\\
-135+135&=0\\
0&\stackrel{\checkmark}=0
\end{align*}
Therefiore, the solution set is $\left\{-3, 0, 3\right\}$.
