Solve 81x^4y^4-16

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

Factor

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

Solution Steps

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

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

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

$$\left(3xy-2\right)\left(3xy+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

$$81\left(xy\right)^{4}-16$$