Solve 1+b+b^2+b^3+b^4+b^5=

$$1 + b + b ^ { 2 } + b ^ { 3 } + b ^ { 4 } + b ^ { 5 } =$$

Factor

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

Solution Steps

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

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

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

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

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

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

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

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

Evaluate

$$b^{5}+b^{4}+b^{3}+b^{2}+b+1$$