Solve 3x^6-81y^3=

$$3 x ^ { 6 } - 81 y ^ { 3 } =$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(x^{6}-27y^{3}\right)$$

Consider x^{6}-27y^{3}. Rewrite x^{6}-27y^{3} as \left(x^{2}\right)^{3}-\left(3y\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).

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

Rewrite the complete factored expression.

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

Evaluate

$$3\left(x^{6}-27y^{3}\right)$$