Answer
0 or $\pm 2$, $\pm 3$, $\pm 5$.
Work Step by Step
$\begin{align}
& {{a}^{5}}\left( {{a}^{2}}-25 \right)+13{{a}^{3}}\left( 25-{{a}^{2}} \right)+36a\left( {{a}^{2}}-25 \right)=0 \\
& {{a}^{5}}\left( {{a}^{2}}-25 \right)-13{{a}^{3}}\left( {{a}^{2}}-25 \right)+36a\left( {{a}^{2}}-25 \right)=0 \\
& \left( {{a}^{2}}-25 \right)\left( {{a}^{5}}-13{{a}^{3}}+36a \right)=0 \\
& a\left( {{a}^{2}}-25 \right)\left( {{a}^{4}}-13{{a}^{2}}+36 \right)=0
\end{align}$
Thus,
$\begin{align}
& \left[ {{\left( {{a}^{2}} \right)}^{2}}-4\left( {{a}^{2}} \right)-9\left( {{a}^{2}} \right)+36 \right]=0 \\
& {{a}^{2}}\left( {{a}^{2}}-4 \right)-9\left( {{a}^{2}}-4 \right)=0 \\
& \left( {{a}^{2}}-4 \right)\left( {{a}^{2}}-9 \right)=0
\end{align}$
So,
$\begin{align}
& \left( {{a}^{2}}-4 \right)=0 \\
& a=\pm 2
\end{align}$
And,
$\begin{align}
& \left( {{a}^{2}}-9 \right)=0 \\
& a=\pm 3
\end{align}$
Therefore, the values of a are $\pm 2$ or $\pm 3$.
Now, take $\left( {{a}^{2}}-25 \right)=0$.
$\begin{align}
& \left( {{a}^{2}}-25 \right)=0 \\
& a=\pm 5
\end{align}$
Thus, the solutions of the expression ${{a}^{5}}\left( {{a}^{2}}-25 \right)+13{{a}^{3}}\left( 25-{{a}^{2}} \right)+36a\left( {{a}^{2}}-25 \right)=0$ are 0, $\pm 2$, $\pm 3$, $\pm 5$.
