We have the recurrence relation:
\begin{align} a_n-c_1a_{n-1}-...-c_ra_{n-r}=0 \tag1\end{align}
Define the polynomial $f$ as:
\begin{align} f(x)=x^n-\sum\limits_{k=1}^{r}c_kx^{n-k} \end{align}
Note that $f(\alpha_*)=0$ because:
\begin{align} f(\alpha_*)=\alpha_*^{n-r}(\alpha_*^r-c_1\alpha_*^{r-1}-...-c_r )=\alpha_*^{n-r}\cdot0=0\end{align}
In fact we know that the multiplicity is 3, so we can factorise the polynomial $f$ by the factor theorem as:
\begin{align} f(x)=(x-\alpha_*)^3g(x) ,\end{align}
where $g(x)$ is a polynomial.
We must show that $a_n=n\alpha_*^n$ and $a_n=n^2\alpha_*^n$ are also solutions of the homogeneous recurrence equation $(1)$
We show that $\alpha_*$ is a root of $f'(x)$, because:
\begin{align} \alpha_*f'(\alpha_*)=\alpha_*\left(n\alpha_*^{n-1}-\sum\limits_{k=1}^{r}c_k(n-k)\alpha_*^{n-k-1}\right)=n\alpha_*^{n}-\sum\limits_{k=1}^{r}c_k(n-k)\alpha_*^{n-k}\end{align}
If we show that the last thing equals zero then it is a solution of the equation $(1)$.
Let us compute $f'(x)$:
\begin{align} f'(x)=3(x-\alpha_*)^2g(x)+(x-\alpha_*)^3g'(x) \end{align}
So:
\begin{align} \alpha_*f'(\alpha_*)=\alpha_*\left(3(\alpha_*-\alpha_*)^2g(\alpha_*)+(\alpha_*-\alpha_*)^3g'(\alpha_*) \right)=0\end{align}
We have shown that $a_n=n\alpha_*^n$ is a solution.
I think you know now what you need to show to prove that $a_n=n^2\alpha_*^n$ is a solution. You just need to do something similar as above. If you really don't know how to proceed a hint is given in the box below.
Show that $\alpha_*$ is a root of $f''(x) $
