Solve (a-b+b/a-b)^2*2a-2b/3a+3b-a^2/a^2-b^2diva/b

$$( \frac { a - b + b } { a - b } ) ^ { 2 } \cdot \frac { 2 a - 2 b } { 3 a + 3 b } - \frac { a ^ { 2 } } { a ^ { 2 } - b ^ { 2 } } \div \frac { a } { b }$$

Evaluate

$$\frac{a\left(2a-3b\right)}{3\left(a^{2}-b^{2}\right)}$$

Short Solution Steps

Combine -b and b to get 0.

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

To raise \frac{a}{a-b} to a power, raise both numerator and denominator to the power and then divide.

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

Multiply \frac{a^{2}}{\left(a-b\right)^{2}} times \frac{2a-2b}{3a+3b} by multiplying numerator times numerator and denominator times denominator.

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

Divide \frac{a^{2}}{a^{2}-b^{2}} by \frac{a}{b} by multiplying \frac{a^{2}}{a^{2}-b^{2}} by the reciprocal of \frac{a}{b}.

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

Cancel out a in both numerator and denominator.

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

Factor \left(a-b\right)^{2}\left(3a+3b\right). Factor a^{2}-b^{2}.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\left(a+b\right)\left(a-b\right)^{2} and \left(a+b\right)\left(a-b\right) is -3\left(a+b\right)\left(a-b\right)^{2}. Multiply \frac{a^{2}\left(2a-2b\right)}{3\left(a+b\right)\left(a-b\right)^{2}} times \frac{-1}{-1}. Multiply \frac{ab}{\left(a+b\right)\left(a-b\right)} times \frac{-3\left(a-b\right)}{-3\left(a-b\right)}.

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

Since \frac{-a^{2}\left(2a-2b\right)}{-3\left(a+b\right)\left(a-b\right)^{2}} and \frac{ab\left(-3\right)\left(a-b\right)}{-3\left(a+b\right)\left(a-b\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in -a^{2}\left(2a-2b\right)-ab\left(-3\right)\left(a-b\right).

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

Combine like terms in -2a^{3}+2a^{2}b+3a^{2}b-3ab^{2}.

$$\frac{-2a^{3}-3ab^{2}+5a^{2}b}{-3\left(a+b\right)\left(a-b\right)^{2}}$$

Expand -3\left(a+b\right)\left(a-b\right)^{2}.

$$\frac{-2a^{3}-3ab^{2}+5a^{2}b}{-3a^{3}+3ab^{2}-3b^{3}+3ba^{2}}$$

Factor the expressions that are not already factored.

$$\frac{a\left(-a+b\right)\left(2a-3b\right)}{3\left(a+b\right)\left(a-b\right)\left(-a+b\right)}$$

Extract the negative sign in -a+b.

$$\frac{-a\left(a-b\right)\left(2a-3b\right)}{3\left(a+b\right)\left(a-b\right)\left(-a+b\right)}$$

Cancel out a-b in both numerator and denominator.

$$\frac{-a\left(2a-3b\right)}{3\left(a+b\right)\left(-a+b\right)}$$

Expand the expression.

$$\frac{-2a^{2}+3ab}{-3a^{2}+3b^{2}}$$

Expand

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

Short Solution Steps

Combine -b and b to get 0.

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

To raise \frac{a}{a-b} to a power, raise both numerator and denominator to the power and then divide.

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

Multiply \frac{a^{2}}{\left(a-b\right)^{2}} times \frac{2a-2b}{3a+3b} by multiplying numerator times numerator and denominator times denominator.

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

Divide \frac{a^{2}}{a^{2}-b^{2}} by \frac{a}{b} by multiplying \frac{a^{2}}{a^{2}-b^{2}} by the reciprocal of \frac{a}{b}.

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

Cancel out a in both numerator and denominator.

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

Factor \left(a-b\right)^{2}\left(3a+3b\right). Factor a^{2}-b^{2}.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\left(a+b\right)\left(a-b\right)^{2} and \left(a+b\right)\left(a-b\right) is -3\left(a+b\right)\left(a-b\right)^{2}. Multiply \frac{a^{2}\left(2a-2b\right)}{3\left(a+b\right)\left(a-b\right)^{2}} times \frac{-1}{-1}. Multiply \frac{ab}{\left(a+b\right)\left(a-b\right)} times \frac{-3\left(a-b\right)}{-3\left(a-b\right)}.

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

Since \frac{-a^{2}\left(2a-2b\right)}{-3\left(a+b\right)\left(a-b\right)^{2}} and \frac{ab\left(-3\right)\left(a-b\right)}{-3\left(a+b\right)\left(a-b\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in -a^{2}\left(2a-2b\right)-ab\left(-3\right)\left(a-b\right).

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

Combine like terms in -2a^{3}+2a^{2}b+3a^{2}b-3ab^{2}.

$$\frac{-2a^{3}-3ab^{2}+5a^{2}b}{-3\left(a+b\right)\left(a-b\right)^{2}}$$

Expand -3\left(a+b\right)\left(a-b\right)^{2}.

$$\frac{-2a^{3}-3ab^{2}+5a^{2}b}{-3a^{3}+3ab^{2}-3b^{3}+3ba^{2}}$$

Factor the expressions that are not already factored.

$$\frac{a\left(-a+b\right)\left(2a-3b\right)}{3\left(a+b\right)\left(a-b\right)\left(-a+b\right)}$$

Extract the negative sign in -a+b.

$$\frac{-a\left(a-b\right)\left(2a-3b\right)}{3\left(a+b\right)\left(a-b\right)\left(-a+b\right)}$$

Cancel out a-b in both numerator and denominator.

$$\frac{-a\left(2a-3b\right)}{3\left(a+b\right)\left(-a+b\right)}$$

Expand the expression.

$$\frac{-2a^{2}+3ab}{-3a^{2}+3b^{2}}$$