Answer
$(t-4)\left(t^2+4 t+4^2\right)$
Work Step by Step
Givenv\begin{equation}
t^3-64.
\end{equation}vWe have a difference of cubes here to factor. This is because $64= 4^3$. Apply the difference of cubes formula: $x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$.
\begin{equation}
\begin{aligned}
t^3-64&= t^3-4^3\\
& =(t-4)\left(t^2+4 t+4^2\right)\\
& =(t-4)\left(t^2+4 t+16\right).
\end{aligned}
\end{equation}The factored form of the difference of cube problem is:
\begin{equation}
t^3-64=(t-4)\left(t^2+4 t+4^2\right).
\end{equation}
