Solve x^8-1^8=

$$x ^ { 8 } - 1 ^ { 8 } =$$

Factor

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

Solution Steps

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

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

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

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

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

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

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

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

Evaluate

$$x^{8}-1$$

Solution Steps

Calculate 1 to the power of 8 and get 1.

$$x^{8}-1$$