Solve 16x^8y^4-81z^4

$$16 x ^ { 8 } y ^ { 4 } - 81 z ^ { 4 }$$

Factor

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

Solution Steps

Rewrite 16x^{8}y^{4}-81z^{4} as \left(4x^{4}y^{2}\right)^{2}-\left(9z^{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).

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

Reorder the terms.

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

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

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

Reorder the terms.

$$\left(2yx^{2}-3z\right)\left(2yx^{2}+3z\right)$$

Rewrite the complete factored expression.

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

Evaluate

$$16y^{4}x^{8}-81z^{4}$$