Solve 1-a^8

$$1 - a ^ { 8 }$$

Factor

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

Solution Steps

Rewrite 1-a^{8} as 1^{2}-\left(-a^{4}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Reorder the terms.

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

Consider -a^{4}+1. Rewrite -a^{4}+1 as 1^{2}-\left(-a^{2}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Reorder the terms.

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

Consider -a^{2}+1. Rewrite -a^{2}+1 as 1^{2}-a^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Reorder the terms.

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: a^{2}+1,a^{4}+1.

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

Evaluate

$$1-a^{8}$$