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

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

Reorder the terms.

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

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

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

Reorder the terms.

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

Rewrite the complete factored expression.

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

Evaluate

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