Solve m^6-1

$$m ^ { 6 } - 1$$

Factor

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

Solution Steps

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

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

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

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

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

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

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

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

Evaluate

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