Solve x^18+x^12+x^6+1

$$x ^ { 18 } + x ^ { 12 } + x ^ { 6 } + 1$$

Factor

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

Solution Steps

Do the grouping x^{18}+x^{12}+x^{6}+1=\left(x^{18}+x^{12}\right)+\left(x^{6}+1\right), and factor out x^{12} in x^{18}+x^{12}.

$$x^{12}\left(x^{6}+1\right)+x^{6}+1$$

Factor out common term x^{6}+1 by using distributive property.

$$\left(x^{6}+1\right)\left(x^{12}+1\right)$$

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).

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

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

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

Evaluate

$$x^{18}+x^{12}+x^{6}+1$$