Solve 64-a^12

$$64- { a }^{ 12 }$$

Factor

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

Solution Steps

Rewrite 64-a^{12} as 8^{2}-\left(-a^{6}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(8+a^{6}\right)\left(8-a^{6}\right)$$

Reorder the terms.

$$\left(a^{6}+8\right)\left(-a^{6}+8\right)$$

Consider a^{6}+8. Rewrite a^{6}+8 as \left(a^{2}\right)^{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).

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

Consider -a^{6}+8. Find one factor of the form ka^{m}+n, where ka^{m} divides the monomial with the highest power -a^{6} and n divides the constant factor 8. One such factor is a^{2}-2. Factor the polynomial by dividing it by this factor.

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: -a^{4}-2a^{2}-4,a^{2}-2,a^{2}+2,a^{4}-2a^{2}+4.

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

Evaluate

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