Solve 512b^18+1=

$$512 b ^ { 18 } + 1 =$$

Factor

$$\left(2b^{2}+1\right)\left(4b^{4}-2b^{2}+1\right)\left(64b^{12}-8b^{6}+1\right)$$

Solution Steps

Rewrite 512b^{18}+1 as \left(8b^{6}\right)^{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).

$$\left(8b^{6}+1\right)\left(64b^{12}-8b^{6}+1\right)$$

Consider 8b^{6}+1. Rewrite 8b^{6}+1 as \left(2b^{2}\right)^{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).

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: 4b^{4}-2b^{2}+1,2b^{2}+1,64b^{12}-8b^{6}+1.

$$\left(4b^{4}-2b^{2}+1\right)\left(2b^{2}+1\right)\left(64b^{12}-8b^{6}+1\right)$$

Evaluate

$$512b^{18}+1$$