Solve a^16-9a^2

$$a ^ { 16 } - 9 a ^ { 2 }$$

Factor

$$a^{2}\left(a^{7}-3\right)\left(a^{7}+3\right)$$

Solution Steps

Factor out a^{2}.

$$a^{2}\left(a^{14}-9\right)$$

Consider a^{14}-9. Rewrite a^{14}-9 as \left(a^{7}\right)^{2}-3^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a^{7}-3\right)\left(a^{7}+3\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: a^{7}-3,a^{7}+3.

$$a^{2}\left(a^{7}-3\right)\left(a^{7}+3\right)$$

Evaluate

$$a^{2}\left(a^{14}-9\right)$$