Solve h^4k^4-1/16

$$h ^ { 4 } k ^ { 4 } - 1 / 16$$

Factor

$$\frac{\left(2hk-1\right)\left(2hk+1\right)\left(4h^{2}k^{2}+1\right)}{16}$$

Solution Steps

Factor out \frac{1}{16}.

$$\frac{16h^{4}k^{4}-1}{16}$$

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

$$\left(4h^{2}k^{2}-1\right)\left(4h^{2}k^{2}+1\right)$$

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

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

Rewrite the complete factored expression.

$$\frac{\left(2hk-1\right)\left(2hk+1\right)\left(4h^{2}k^{2}+1\right)}{16}$$

Evaluate

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