Rewrite $64-a^{6}$ as $8^{2}-\left(-a^{3}\right)^{2}$. The difference of squares can be factored using the rule: $p^{2}-q^{2}=\left(p-q\right)\left(p+q\right)$.
$$\left(8+a^{3}\right)\left(8-a^{3}\right)$$
Reorder the terms.
$$\left(a^{3}+8\right)\left(-a^{3}+8\right)$$
Consider $a^{3}+8$. Rewrite $a^{3}+8$ as $a^{3}+2^{3}$. The sum of cubes can be factored using the rule: $p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right)$.
$$\left(a+2\right)\left(a^{2}-2a+4\right)$$
Consider $-a^{3}+8$. By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $8$ and $q$ divides the leading coefficient $-1$. One such root is $2$. Factor the polynomial by dividing it by $a-2$.
$$\left(a-2\right)\left(-a^{2}-2a-4\right)$$
Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: $-a^{2}-2a-4,a^{2}-2a+4$.
