Solve x^16-1

$$x ^ { 16 } - 1$$

Factor

$$\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right)$$

Solution Steps

Rewrite x^{16}-1 as \left(x^{8}\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).

$$\left(x^{8}-1\right)\left(x^{8}+1\right)$$

Consider x^{8}-1. 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,x^{8}+1.

$$\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right)$$

Evaluate

$$x^{16}-1$$