Consider $64a^{3}-1$. Rewrite $64a^{3}-1$ as $\left(4a\right)^{3}-1^{3}$. The difference of cubes can be factored using the rule: $p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right)$.
$$\left(4a-1\right)\left(16a^{2}+4a+1\right)$$
Rewrite the complete factored expression. Polynomial $16a^{2}+4a+1$ is not factored since it does not have any rational roots.
