Solve 27x^3+8y^6x^9

$$27 x ^ { 3 } + 8 y ^ { 6 } x ^ { 9 }$$

Factor

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

Solution Steps

Factor out x^{3}.

$$x^{3}\left(27+8y^{6}x^{6}\right)$$

Consider 27+8y^{6}x^{6}. Rewrite 27+8y^{6}x^{6} as \left(2x^{2}y^{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).

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

Rewrite the complete factored expression.

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

Evaluate

$$x^{3}\left(8\left(xy\right)^{6}+27\right)$$