Solve x^3-8/27y^6

$$x ^ { 3 } - \frac { 8 } { 27 } y ^ { 6 }$$

Factor

$$\frac{\left(3x-2y^{2}\right)\left(9x^{2}+6xy^{2}+4y^{4}\right)}{27}$$

Solution Steps

Factor out \frac{1}{27}.

$$\frac{27x^{3}-8y^{6}}{27}$$

Consider 27x^{3}-8y^{6}. Rewrite 27x^{3}-8y^{6} as \left(3x\right)^{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).

$$\left(3x-2y^{2}\right)\left(9x^{2}+6xy^{2}+4y^{4}\right)$$

Rewrite the complete factored expression.

$$\frac{\left(3x-2y^{2}\right)\left(9x^{2}+6xy^{2}+4y^{4}\right)}{27}$$

Evaluate

$$-\frac{8y^{6}}{27}+x^{3}$$