Solve x^100+x^10

$${ x }^{ 100 } + { x }^{ 10 }$$

Factor

$$\left(x^{2}+1\right)\left(x^{8}-x^{6}+x^{4}-x^{2}+1\right)x^{10}\left(x^{20}-x^{10}+1\right)\left(x^{60}-x^{30}+1\right)$$

Solution Steps

Factor out x^{10}.

$$x^{10}\left(x^{90}+1\right)$$

Consider x^{90}+1. Rewrite x^{90}+1 as \left(x^{30}\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).

$$\left(x^{30}+1\right)\left(x^{60}-x^{30}+1\right)$$

Consider x^{30}+1. Rewrite x^{30}+1 as \left(x^{10}\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).

$$\left(x^{10}+1\right)\left(x^{20}-x^{10}+1\right)$$

Consider x^{10}+1. Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{10} and m divides the constant factor 1. One such factor is x^{2}+1. Factor the polynomial by dividing it by this factor.

$$\left(x^{2}+1\right)\left(x^{8}-x^{6}+x^{4}-x^{2}+1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{8}-x^{6}+x^{4}-x^{2}+1,x^{2}+1,x^{20}-x^{10}+1,x^{60}-x^{30}+1.

$$x^{10}\left(x^{2}+1\right)\left(x^{8}-x^{6}+x^{4}-x^{2}+1\right)\left(x^{20}-x^{10}+1\right)\left(x^{60}-x^{30}+1\right)$$

Evaluate

$$x^{100}+x^{10}$$