We have $a_n=-\left(\frac{n^3+5}{n^2-8n}\right).$ Informally, we now want to show that that if $n$ is "large" then $\dfrac{n^3+5}{n^2-8n}$ is large. We do not want to worry about negative values of $n^2-8n$, so suppose that $n\gt 8$, making $n^2-8n\gt 0$. For $n \gt 8$, we have $\frac{n^3+5}{n^2-8n} \gt \frac{n^3}{n^2-8n},\tag{$1$}$ for on the right we have made the numerator smaller than on the left.
Also, for $n\gt 8$, we have $\frac{n^3}{n^2-8n}\gt \frac{n^3}{n^2},\tag{$2$}$ for on the right we have made the denominator bigger than on the left.
Combining Inequalities $(1)$ and $(2)$, we get that if $n\gt 8$ then $\frac{n^3+5}{n^2-8}\gt \frac{n^3}{n^2}=n,$ and therefore $a_n=-\left(\frac{n^3+5}{n^2-8}\right)\lt -n.\tag{$3$}$
Finally, we want to show that for any number $K$, so matter how large negative it is, we can find $N$ such that if $n\gt N$, then $a_n\lt K$. That will show that $\lim_{n\to\infty}a_n=-\infty$.
So let $K$ be negative, and let $N$ be any integer greater than the larger of $8$ and $|K|$. By Inequality $(3)$, if $n\gt N$, we have $a_n\lt -n \lt -N\lt -|K|=K.$
Remark: It is useful to work as long as possible with positive quantities, since there our intuition is likely to be more accurate.
There were additional items of unpleasantness introduced because of the fact that the limit is $-\infty$. It would be more comfortable to let $b_n=-a_n$, and show that $\lim_{n\to\infty} b_n=\infty$, and then refer to a little theorem somewhere to the effect that a sequence $(x_n)$ has limit $p$ (which may be a real number or one of $\pm\infty$) iff the sequence $(-x_n)$ has limit $-p$.