Solve x^96-1

$${ x }^{ 96 } -1$$

Factor

$$\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right)\left(x^{16}+1\right)\left(x^{60}-x^{58}+x^{54}-x^{52}+x^{48}-x^{46}+x^{42}-x^{40}+x^{36}-x^{34}+x^{30}-x^{26}+x^{24}-x^{20}+x^{18}-x^{14}+x^{12}-x^{8}+x^{6}-x^{2}+1\right)$$

Solution Steps

Rewrite x^{96}-1 as \left(x^{48}\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).

$$\left(x^{48}-1\right)\left(x^{48}+1\right)$$

Consider x^{48}-1. Rewrite x^{48}-1 as \left(x^{24}\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).

$$\left(x^{24}-1\right)\left(x^{24}+1\right)$$

Consider x^{24}-1. Rewrite x^{24}-1 as \left(x^{12}\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).

$$\left(x^{12}-1\right)\left(x^{12}+1\right)$$

Consider x^{12}-1. Rewrite x^{12}-1 as \left(x^{6}\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).

$$\left(x^{6}-1\right)\left(x^{6}+1\right)$$

Consider x^{6}-1. Rewrite x^{6}-1 as \left(x^{3}\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).

$$\left(x^{3}-1\right)\left(x^{3}+1\right)$$

Consider x^{3}-1. Rewrite x^{3}-1 as x^{3}-1^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

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

Consider x^{3}+1. Rewrite x^{3}+1 as x^{3}+1^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Consider x^{6}+1. Rewrite x^{6}+1 as \left(x^{2}\right)^{3}+1^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Consider x^{12}+1. Rewrite x^{12}+1 as \left(x^{4}\right)^{3}+1^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Consider x^{24}+1. Rewrite x^{24}+1 as \left(x^{8}\right)^{3}+1^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

$$\left(x^{8}+1\right)\left(x^{16}-x^{8}+1\right)$$

Consider x^{48}+1. Rewrite x^{48}+1 as \left(x^{16}\right)^{3}+1^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

$$\left(x^{16}+1\right)\left(x^{32}-x^{16}+1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}-x+1,x^{2}+x+1,x^{4}-x^{2}+1,x^{2}+1,x^{8}-x^{4}+1,x^{4}+1,x^{16}-x^{8}+1,x^{8}+1,x^{32}-x^{16}+1,x^{16}+1.

$$\left(x-1\right)\left(x^{2}-x+1\right)\left(x+1\right)\left(x^{2}+x+1\right)\left(x^{4}-x^{2}+1\right)\left(x^{2}+1\right)\left(x^{8}-x^{4}+1\right)\left(x^{4}+1\right)\left(x^{16}-x^{8}+1\right)\left(x^{8}+1\right)\left(x^{32}-x^{16}+1\right)\left(x^{16}+1\right)$$

Evaluate

$$x^{96}-1$$