Solve 3x^3y^6+81=

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

Factor

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

Solution Steps

Factor out 3.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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