Solve 64m^6y^3+8m^3n^6=

$$64 m ^ { 6 } y ^ { 3 } + 8 m ^ { 3 } n ^ { 6 } =$$

Factor

$$8\left(2my+n^{2}\right)m^{3}\left(4m^{2}y^{2}-2myn^{2}+n^{4}\right)$$

Solution Steps

Factor out 8.

$$8\left(8m^{6}y^{3}+m^{3}n^{6}\right)$$

Consider 8m^{6}y^{3}+m^{3}n^{6}. Factor out m^{3}.

$$m^{3}\left(8m^{3}y^{3}+n^{6}\right)$$

Consider 8m^{3}y^{3}+n^{6}. Rewrite 8m^{3}y^{3}+n^{6} as \left(2my\right)^{3}+\left(n^{2}\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).

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

Rewrite the complete factored expression.

$$8m^{3}\left(2my+n^{2}\right)\left(4m^{2}y^{2}-2myn^{2}+n^{4}\right)$$

Evaluate

$$8m^{3}\left(8\left(my\right)^{3}+n^{6}\right)$$