Solve (16x^8y^4/81-y^12)

$$( \frac{ 16 { x }^{ 8 } { y }^{ 4 } }{ 81 } - { y }^{ 12 } )$$

Factor

$$\frac{\left(2x^{2}-3y^{2}\right)\left(2x^{2}+3y^{2}\right)y^{4}\left(4x^{4}+9y^{4}\right)}{81}$$

Solution Steps

Factor out \frac{1}{81}.

$$\frac{16x^{8}y^{4}-81y^{12}}{81}$$

Consider 16x^{8}y^{4}-81y^{12}. Factor out y^{4}.

$$y^{4}\left(16x^{8}-81y^{8}\right)$$

Consider 16x^{8}-81y^{8}. Rewrite 16x^{8}-81y^{8} as \left(4x^{4}\right)^{2}-\left(9y^{4}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider 4x^{4}-9y^{4}. Rewrite 4x^{4}-9y^{4} as \left(2x^{2}\right)^{2}-\left(3y^{2}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Rewrite the complete factored expression.

$$\frac{y^{4}\left(2x^{2}-3y^{2}\right)\left(2x^{2}+3y^{2}\right)\left(4x^{4}+9y^{4}\right)}{81}$$

Evaluate

$$\frac{16y^{4}x^{8}}{81}-y^{12}$$