Solve m^8-n^8

$$m ^ { 8 } - n ^ { 8 }$$

Factor

$$\left(m-n\right)\left(m+n\right)\left(m^{2}+n^{2}\right)\left(m^{4}+n^{4}\right)$$

Solution Steps

Rewrite m^{8}-n^{8} as \left(m^{4}\right)^{2}-\left(n^{4}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(m^{4}-n^{4}\right)\left(m^{4}+n^{4}\right)$$

Consider m^{4}-n^{4}. Rewrite m^{4}-n^{4} as \left(m^{2}\right)^{2}-\left(n^{2}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider m^{2}-n^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(m-n\right)\left(m+n\right)$$

Rewrite the complete factored expression.

$$\left(m-n\right)\left(m+n\right)\left(m^{2}+n^{2}\right)\left(m^{4}+n^{4}\right)$$

Evaluate

$$m^{8}-n^{8}$$