Solve x+x^2+x^3+x^4+x^5+{x}^6

$$x+ { x }^{ 2 } + { x }^{ 3 } + { x }^{ 4 } + { x }^{ 5 } + { x }^{ 6 }$$

Factor

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

Solution Steps

Factor out x.

$$x\left(1+x+x^{2}+x^{3}+x^{4}+x^{5}\right)$$

Consider 1+x+x^{2}+x^{3}+x^{4}+x^{5}. Do the grouping 1+x+x^{2}+x^{3}+x^{4}+x^{5}=\left(1+x+x^{2}\right)+\left(x^{3}+x^{4}+x^{5}\right), and factor out x^{3} in x^{3}+x^{4}+x^{5}.

$$x^{2}+x+1+x^{3}\left(x^{2}+x+1\right)$$

Factor out common term x^{2}+x+1 by using distributive property.

$$\left(x^{2}+x+1\right)\left(1+x^{3}\right)$$

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.

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

Evaluate

$$x\left(x^{5}+x^{4}+x^{3}+x^{2}+x+1\right)$$