Solve (a-2b/a^{2+2ab}-1/a^{2-4b^{2}}:a+2b/(2b-a)^2):a^2-2ab/a^2+4ab+4b^2=

$$( \frac { a - 2 b } { a ^ { 2 } + 2 a b } - \frac { 1 } { a ^ { 2 } - 4 b ^ { 2 } } : \frac { a + 2 b } { ( 2 b - a ) ^ { 2 } } ) : \frac { a ^ { 2 } - 2 a b } { a ^ { 2 } + 4 a b + 4 b ^ { 2 } } =$$

Evaluate

$$\frac{2b}{a^{2}}$$

Short Solution Steps

Divide \frac{1}{a^{2}-4b^{2}} by \frac{a+2b}{\left(2b-a\right)^{2}} by multiplying \frac{1}{a^{2}-4b^{2}} by the reciprocal of \frac{a+2b}{\left(2b-a\right)^{2}}.

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

Factor a^{2}+2ab. Factor \left(a^{2}-4b^{2}\right)\left(a+2b\right).

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a\left(a+2b\right) and \left(a-2b\right)\left(a+2b\right)^{2} is a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}. Multiply \frac{a-2b}{a\left(a+2b\right)} times \frac{\left(a-2b\right)\left(a+2b\right)^{2}}{\left(a-2b\right)\left(a+2b\right)^{2}}. Multiply \frac{\left(2b-a\right)^{2}}{\left(a-2b\right)\left(a+2b\right)^{2}} times \frac{a\left(a+2b\right)}{a\left(a+2b\right)}.

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

Since \frac{\left(a-2b\right)\left(a-2b\right)\left(a+2b\right)^{2}}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}} and \frac{\left(2b-a\right)^{2}a\left(a+2b\right)}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(a-2b\right)\left(a-2b\right)\left(a+2b\right)^{2}-\left(2b-a\right)^{2}a\left(a+2b\right).

$$\frac{\frac{a^{4}-8ab^{3}+2a^{3}b-4a^{2}b^{2}+8ab^{3}-2a^{3}b-4a^{2}b^{2}+16b^{4}-4b^{2}a^{2}-8b^{3}a+4ba^{3}+8b^{2}a^{2}-a^{4}-2a^{3}b}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}}}{\frac{a^{2}-2ab}{a^{2}+4ab+4b^{2}}}$$

Combine like terms in a^{4}-8ab^{3}+2a^{3}b-4a^{2}b^{2}+8ab^{3}-2a^{3}b-4a^{2}b^{2}+16b^{4}-4b^{2}a^{2}-8b^{3}a+4ba^{3}+8b^{2}a^{2}-a^{4}-2a^{3}b.

$$\frac{\frac{2a^{3}b-8ab^{3}-4a^{2}b^{2}+16b^{4}}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}}}{\frac{a^{2}-2ab}{a^{2}+4ab+4b^{2}}}$$

Factor the expressions that are not already factored in \frac{2a^{3}b-8ab^{3}-4a^{2}b^{2}+16b^{4}}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}}.

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

Cancel out \left(a-2b\right)\left(a+2b\right) in both numerator and denominator.

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

Divide \frac{2b\left(a-2b\right)}{a\left(a+2b\right)^{2}} by \frac{a^{2}-2ab}{a^{2}+4ab+4b^{2}} by multiplying \frac{2b\left(a-2b\right)}{a\left(a+2b\right)^{2}} by the reciprocal of \frac{a^{2}-2ab}{a^{2}+4ab+4b^{2}}.

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

Factor the expressions that are not already factored.

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

Cancel out \left(a-2b\right)\left(a+2b\right)^{2} in both numerator and denominator.

$$\frac{2b}{a^{2}}$$

Expand

$$\frac{2b}{a^{2}}$$

Short Solution Steps

Divide \frac{1}{a^{2}-4b^{2}} by \frac{a+2b}{\left(2b-a\right)^{2}} by multiplying \frac{1}{a^{2}-4b^{2}} by the reciprocal of \frac{a+2b}{\left(2b-a\right)^{2}}.

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

Factor a^{2}+2ab. Factor \left(a^{2}-4b^{2}\right)\left(a+2b\right).

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a\left(a+2b\right) and \left(a-2b\right)\left(a+2b\right)^{2} is a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}. Multiply \frac{a-2b}{a\left(a+2b\right)} times \frac{\left(a-2b\right)\left(a+2b\right)^{2}}{\left(a-2b\right)\left(a+2b\right)^{2}}. Multiply \frac{\left(2b-a\right)^{2}}{\left(a-2b\right)\left(a+2b\right)^{2}} times \frac{a\left(a+2b\right)}{a\left(a+2b\right)}.

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

Since \frac{\left(a-2b\right)\left(a-2b\right)\left(a+2b\right)^{2}}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}} and \frac{\left(2b-a\right)^{2}a\left(a+2b\right)}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(a-2b\right)\left(a-2b\right)\left(a+2b\right)^{2}-\left(2b-a\right)^{2}a\left(a+2b\right).

$$\frac{\frac{a^{4}-8ab^{3}+2a^{3}b-4a^{2}b^{2}+8ab^{3}-2a^{3}b-4a^{2}b^{2}+16b^{4}-4b^{2}a^{2}-8b^{3}a+4ba^{3}+8b^{2}a^{2}-a^{4}-2a^{3}b}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}}}{\frac{a^{2}-2ab}{a^{2}+4ab+4b^{2}}}$$

Combine like terms in a^{4}-8ab^{3}+2a^{3}b-4a^{2}b^{2}+8ab^{3}-2a^{3}b-4a^{2}b^{2}+16b^{4}-4b^{2}a^{2}-8b^{3}a+4ba^{3}+8b^{2}a^{2}-a^{4}-2a^{3}b.

$$\frac{\frac{2a^{3}b-8ab^{3}-4a^{2}b^{2}+16b^{4}}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}}}{\frac{a^{2}-2ab}{a^{2}+4ab+4b^{2}}}$$

Factor the expressions that are not already factored in \frac{2a^{3}b-8ab^{3}-4a^{2}b^{2}+16b^{4}}{a\left(a-2b\right)\left(a+2b\right)\left(a+2b\right)^{2}}.

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

Cancel out \left(a-2b\right)\left(a+2b\right) in both numerator and denominator.

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

Divide \frac{2b\left(a-2b\right)}{a\left(a+2b\right)^{2}} by \frac{a^{2}-2ab}{a^{2}+4ab+4b^{2}} by multiplying \frac{2b\left(a-2b\right)}{a\left(a+2b\right)^{2}} by the reciprocal of \frac{a^{2}-2ab}{a^{2}+4ab+4b^{2}}.

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

Factor the expressions that are not already factored.

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

Cancel out \left(a-2b\right)\left(a+2b\right)^{2} in both numerator and denominator.

$$\frac{2b}{a^{2}}$$