Solve a^6-a^4+a^2-1=

$$a ^ { 6 } - a ^ { 4 } + a ^ { 2 } - 1 =$$

Factor

$$\left(a-1\right)\left(a+1\right)\left(a^{4}+1\right)$$

Solution Steps

Do the grouping a^{6}-a^{4}+a^{2}-1=\left(a^{6}-a^{4}\right)+\left(a^{2}-1\right), and factor out a^{4} in a^{6}-a^{4}.

$$a^{4}\left(a^{2}-1\right)+a^{2}-1$$

Factor out common term a^{2}-1 by using distributive property.

$$\left(a^{2}-1\right)\left(a^{4}+1\right)$$

Consider a^{2}-1. Rewrite a^{2}-1 as a^{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).

$$\left(a-1\right)\left(a+1\right)$$

Rewrite the complete factored expression. Polynomial a^{4}+1 is not factored since it does not have any rational roots.

$$\left(a-1\right)\left(a+1\right)\left(a^{4}+1\right)$$

Evaluate

$$\left(a^{2}-1\right)\left(a^{4}+1\right)$$