Solve 64-a^6

$$64 - a ^ { 6 }$$

Factor

$$\left(a-2\right)\left(a+2\right)\left(-a^{2}+2a-4\right)\left(a^{2}+2a+4\right)$$

Solution Steps

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.

$$\left(-a^{2}-2a-4\right)\left(a-2\right)\left(a+2\right)\left(a^{2}-2a+4\right)$$

Evaluate

$$\left(4-a^{2}\right)\left(\left(a^{2}+4\right)^{2}-4a^{2}\right)$$