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