Solve 25a^6b^4-1/4

$$25 a ^ { 6 } b ^ { 4 } - \frac { 1 } { 4 }$$

Factor

$$\frac{\left(10b^{2}a^{3}-1\right)\left(10b^{2}a^{3}+1\right)}{4}$$

Solution Steps

Factor out \frac{1}{4}.

$$\frac{100a^{6}b^{4}-1}{4}$$

Consider 100a^{6}b^{4}-1. Rewrite 100a^{6}b^{4}-1 as \left(10a^{3}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).

$$\left(10a^{3}b^{2}-1\right)\left(10a^{3}b^{2}+1\right)$$

Reorder the terms.

$$\left(10b^{2}a^{3}-1\right)\left(10b^{2}a^{3}+1\right)$$

Rewrite the complete factored expression.

$$\frac{\left(10b^{2}a^{3}-1\right)\left(10b^{2}a^{3}+1\right)}{4}$$

Evaluate

$$25b^{4}a^{6}-\frac{1}{4}$$