Answer
$$A^3= \left[\begin{array}{ll}{27} & {0}\\
{-49}&{125}\end{array}\right]$$
Work Step by Step
Let $$
A=\left[\begin{array}{ll}{3} & {0}\\
{-1}&{5}\end{array}\right],
$$
we have
$$A^3=AAA=\left[\begin{array}{ll}{3} & {0}\\
{-1}&{5}\end{array}\right]\left[\begin{array}{ll}{3} & {0}\\
{-1}&{5}\end{array}\right]\left[\begin{array}{ll}{3} & {0}\\
{-1}&{5}\end{array}\right]\\
=\left[\begin{array}{ll}{3} & {0}\\
{-1}&{5}\end{array}\right]\left[\begin{array}{ll}{9} & {0}\\
{-8}&{25}\end{array}\right]=\left[\begin{array}{ll}{27} & {0}\\
{-49}&{125}\end{array}\right].$$