Solve 4p^6q-4q

$$4 p ^ { 6 } q - 4 q$$

Factor

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

Solution Steps

Factor out 4.

$$4\left(p^{6}q-q\right)$$

Consider p^{6}q-q. Factor out q.

$$q\left(p^{6}-1\right)$$

Consider p^{6}-1. Rewrite p^{6}-1 as \left(p^{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).

$$\left(p^{3}-1\right)\left(p^{3}+1\right)$$

Consider p^{3}-1. Rewrite p^{3}-1 as p^{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).

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

Consider p^{3}+1. Rewrite p^{3}+1 as p^{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).

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

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

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

Evaluate

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