Solve m^12n^6-m^3n^15=

$$m ^ { 12 } n ^ { 6 } - m ^ { 3 } n ^ { 15 } =$$

Factor

$$\left(m-n\right)m^{3}\left(m^{2}+mn+n^{2}\right)n^{6}\left(m^{6}+m^{3}n^{3}+n^{6}\right)$$

Solution Steps

Factor out m^{3}n^{6}.

$$m^{3}n^{6}\left(m^{9}-n^{9}\right)$$

Consider m^{9}-n^{9}. Rewrite m^{9}-n^{9} as \left(m^{3}\right)^{3}-\left(n^{3}\right)^{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).

$$\left(m^{3}-n^{3}\right)\left(m^{6}+m^{3}n^{3}+n^{6}\right)$$

Consider m^{3}-n^{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).

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

Rewrite the complete factored expression.

$$m^{3}n^{6}\left(m-n\right)\left(m^{2}+mn+n^{2}\right)\left(m^{6}+m^{3}n^{3}+n^{6}\right)$$

Evaluate

$$m^{3}n^{6}\left(m^{9}-n^{9}\right)$$