Rewrite $a^{6}-1$ as $\left(a^{3}\right)^{2}-1^{2}$. The difference of squares can be factored using the rule: $p^{2}-q^{2}=\left(p-q\right)\left(p+q\right)$.
$$\left(a^{3}-1\right)\left(a^{3}+1\right)$$
Consider $a^{3}-1$. Rewrite $a^{3}-1$ as $a^{3}-1^{3}$. The difference of cubes can be factored using the rule: $p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right)$.
$$\left(a-1\right)\left(a^{2}+a+1\right)$$
Consider $a^{3}+1$. Rewrite $a^{3}+1$ as $a^{3}+1^{3}$. The sum of cubes can be factored using the rule: $p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right)$.
$$\left(a+1\right)\left(a^{2}-a+1\right)$$
Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: $a^{2}-a+1,a^{2}+a+1$.
