Solve 16a^4b^4-1

$$16 a ^ { 4 } b ^ { 4 } - 1$$

Factor

$$\left(2ab-1\right)\left(2ab+1\right)\left(4a^{2}b^{2}+1\right)$$

Solution Steps

Rewrite 16a^{4}b^{4}-1 as \left(4a^{2}b^{2}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(4a^{2}b^{2}-1\right)\left(4a^{2}b^{2}+1\right)$$

Consider 4a^{2}b^{2}-1. Rewrite 4a^{2}b^{2}-1 as \left(2ab\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(2ab-1\right)\left(2ab+1\right)$$

Rewrite the complete factored expression.

$$\left(2ab-1\right)\left(2ab+1\right)\left(4a^{2}b^{2}+1\right)$$

Evaluate

$$16\left(ab\right)^{4}-1$$