Solve a^8-a^2

$$a ^ { 8 } - a ^ { 2 }$$

Factor

$$\left(a-1\right)\left(a+1\right)a^{2}\left(a^{2}-a+1\right)\left(a^{2}+a+1\right)$$

Solution Steps

Factor out a^{2}.

$$a^{2}\left(a^{6}-1\right)$$

Consider a^{6}-1. 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.

$$a^{2}\left(a-1\right)\left(a^{2}+a+1\right)\left(a+1\right)\left(a^{2}-a+1\right)$$

Evaluate

$$a^{2}\left(a^{2}-1\right)\left(\left(a^{2}+1\right)^{2}-a^{2}\right)$$