Answer
$y=x+5$ is Oblique asymptote.
$x=3$ is a vertical asymptote
Work Step by Step
$H(x)=\frac{x^3-8}{x^2-5x+6}=\frac{(x-2)(x^2+2x+4)}{(x-2)(x-3)}=\frac{x^2+2x+4}{x-3},x\not=2,x\not=3$
The degree of the leading coefficient of the numerator is, $n=2$. the degree of the leading coefficient the denominator is, $m=1$.
Thus, $n=m+1,$
When $n=m+1,$ the quotient of the division of numerator by denominator is Oblique asymptote.
$\begin{array} x &x+5\\
&-- -- -- --\\
x^2-5x+6 |& x^3-8\\
& -x^3+5x^2-6x\\
& -- -- -- -- \\
& 5x^2-6x-8\\
& -5x^2+25x-30\\
&-- -- -- -- \\
& 19x-38
\end{array}$
Thus, $y=x+5$ is Oblique asymptote.
Factorizing the denominator,
$x^2-5x+6=x^2-2x-3x+6,$
$=x(x-2)-3(x-2),$
$=(x-3)(x-2),$
Thus, $x\in \{2, 3\}$ is a vertical asymptote
