Answer
V.A. $x=3$, O.A. $y=x+5$.
Work Step by Step
Given $y=\frac{x^3-8}{x^2-5x+6}=\frac{(x-2)(x^2+2x+4)}{(x-2)(x-3)}=\frac{x^2+2x+4}{x-3}=\frac{x^2-3x+5x-15+19}{x-3}=x+5+\frac{15}{x-3}, x\ne2$, we can find its vertical asymptote(s) V.A. $x=3$, horizontal asymptote(s) H.A. $none$, oblique asymptote(s) O.A. $y=x+5$.
