Solve x^10-x^6-x^4+1

$$x ^ { 10 } - x ^ { 6 } - x ^ { 4 } + 1$$

Factor

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

Solution Steps

Do the grouping x^{10}-x^{6}-x^{4}+1=\left(x^{10}-x^{6}\right)+\left(-x^{4}+1\right), and factor out x^{6} in the first and -1 in the second group.

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

Factor out common term x^{4}-1 by using distributive property.

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

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).

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

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).

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

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^{2}+1.

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

Evaluate

$$x^{10}-x^{6}-x^{4}+1$$