Solve (a+b/a-b-a-b/a+b)*a^2-b^2/16ab

$$( \frac { a + b } { a - b } - \frac { a - b } { a + b } ) \cdot \frac { a ^ { 2 } - b ^ { 2 } } { 16 a b }$$

Evaluate

$$\frac{1}{4}=0.25$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a-b and a+b is \left(a+b\right)\left(a-b\right). Multiply \frac{a+b}{a-b} times \frac{a+b}{a+b}. Multiply \frac{a-b}{a+b} times \frac{a-b}{a-b}.

$$\left(\frac{\left(a+b\right)\left(a+b\right)}{\left(a+b\right)\left(a-b\right)}-\frac{\left(a-b\right)\left(a-b\right)}{\left(a+b\right)\left(a-b\right)}\right)\times \frac{a^{2}-b^{2}}{16ab}$$

Since \frac{\left(a+b\right)\left(a+b\right)}{\left(a+b\right)\left(a-b\right)} and \frac{\left(a-b\right)\left(a-b\right)}{\left(a+b\right)\left(a-b\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(a+b\right)\left(a+b\right)-\left(a-b\right)\left(a-b\right)}{\left(a+b\right)\left(a-b\right)}\times \frac{a^{2}-b^{2}}{16ab}$$

Do the multiplications in \left(a+b\right)\left(a+b\right)-\left(a-b\right)\left(a-b\right).

$$\frac{a^{2}+ab+ab+b^{2}-a^{2}+ab+ab-b^{2}}{\left(a+b\right)\left(a-b\right)}\times \frac{a^{2}-b^{2}}{16ab}$$

Combine like terms in a^{2}+ab+ab+b^{2}-a^{2}+ab+ab-b^{2}.

$$\frac{4ab}{\left(a+b\right)\left(a-b\right)}\times \frac{a^{2}-b^{2}}{16ab}$$

Multiply \frac{4ab}{\left(a+b\right)\left(a-b\right)} times \frac{a^{2}-b^{2}}{16ab} by multiplying numerator times numerator and denominator times denominator.

$$\frac{4ab\left(a^{2}-b^{2}\right)}{\left(a+b\right)\left(a-b\right)\times 16ab}$$

Cancel out 4ab in both numerator and denominator.

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

Factor the expressions that are not already factored.

$$\frac{\left(a+b\right)\left(a-b\right)}{4\left(a+b\right)\left(a-b\right)}$$

Cancel out \left(a+b\right)\left(a-b\right) in both numerator and denominator.

$$\frac{1}{4}$$

Factor

$$\frac{1}{2 ^ {2}} = 0.25$$