Solve 4mx^3-32my^6=

$$4 m x ^ { 3 } - 32 m y ^ { 6 } =$$

Factor

$$4m\left(x-2y^{2}\right)\left(x^{2}+2xy^{2}+4y^{4}\right)$$

Solution Steps

Factor out 4.

$$4\left(mx^{3}-8my^{6}\right)$$

Consider mx^{3}-8my^{6}. Factor out m.

$$m\left(x^{3}-8y^{6}\right)$$

Consider x^{3}-8y^{6}. Rewrite x^{3}-8y^{6} as x^{3}-\left(2y^{2}\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).

$$\left(x-2y^{2}\right)\left(x^{2}+2xy^{2}+4y^{4}\right)$$

Rewrite the complete factored expression.

$$4m\left(x-2y^{2}\right)\left(x^{2}+2xy^{2}+4y^{4}\right)$$

Evaluate

$$4m\left(x^{3}-8y^{6}\right)$$