Solve 64m^6+n^12

$$64 m ^ { 6 } + n ^ { 12 }$$

Factor

$$\left(4m^{2}+n^{4}\right)\left(16m^{4}-4m^{2}n^{4}+n^{8}\right)$$

Solution Steps

Rewrite 64m^{6}+n^{12} as \left(4m^{2}\right)^{3}+\left(n^{4}\right)^{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).

$$\left(4m^{2}+n^{4}\right)\left(16m^{4}-4m^{2}n^{4}+n^{8}\right)$$

Evaluate

$$64m^{6}+n^{12}$$