Consider $8a^{3}+27$. Rewrite $8a^{3}+27$ as $\left(2a\right)^{3}+3^{3}$. The sum of cubes can be factored using the rule: $p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right)$.
$$\left(2a+3\right)\left(4a^{2}-6a+9\right)$$
Rewrite the complete factored expression. Polynomial $4a^{2}-6a+9$ is not factored since it does not have any rational roots.
