Solve {l}{frac{a^{2}-b^2}{a+b}*ab+a/a^2-2ab+b^2*a^2-ab/2b+2}{-b)quada(1-b}

$$\left. \begin{array} { l } { \frac { a ^ { 2 } - b ^ { 2 } } { a + b } \cdot \frac { a b + a } { a ^ { 2 } - 2 a b + b ^ { 2 } } \cdot \frac { a ^ { 2 } - a b } { 2 b + 2 } } \\ { - b ) \quad a ( 1 - b } \end{array} \right.$$

Evaluate

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{a^{2}-b^{2}}{a+b}.

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

Cancel out a+b in both numerator and denominator.

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

Express \left(a-b\right)\times \frac{ab+a}{a^{2}-2ab+b^{2}} as a single fraction.

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

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

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

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

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

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

$$\frac{\left(a-b\right)\left(ab+a\right)\left(a^{2}-ab\right)}{2\left(b+1\right)\left(a-b\right)^{2}}-\frac{b\times 2\left(b+1\right)\left(a-b\right)^{2}}{2\left(b+1\right)\left(a-b\right)^{2}}$$

Since \frac{\left(a-b\right)\left(ab+a\right)\left(a^{2}-ab\right)}{2\left(b+1\right)\left(a-b\right)^{2}} and \frac{b\times 2\left(b+1\right)\left(a-b\right)^{2}}{2\left(b+1\right)\left(a-b\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(a-b\right)\left(ab+a\right)\left(a^{2}-ab\right)-b\times 2\left(b+1\right)\left(a-b\right)^{2}}{2\left(b+1\right)\left(a-b\right)^{2}}$$

Do the multiplications in \left(a-b\right)\left(ab+a\right)\left(a^{2}-ab\right)-b\times 2\left(b+1\right)\left(a-b\right)^{2}.

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

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

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

Factor the expressions that are not already factored in \frac{a^{4}b-2b^{3}+b^{3}a^{2}-2a^{3}b^{2}+a^{4}+4b^{3}a-2a^{3}b-b^{2}a^{2}-2b^{4}+4b^{2}a-2ba^{2}}{2\left(b+1\right)\left(a-b\right)^{2}}.

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

Cancel out \left(b+1\right)\left(a-b\right)^{2} in both numerator and denominator.

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

Expand

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{a^{2}-b^{2}}{a+b}.

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

Cancel out a+b in both numerator and denominator.

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

Express \left(a-b\right)\times \frac{ab+a}{a^{2}-2ab+b^{2}} as a single fraction.

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

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

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

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

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

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

$$\frac{\left(a-b\right)\left(ab+a\right)\left(a^{2}-ab\right)}{2\left(b+1\right)\left(a-b\right)^{2}}-\frac{b\times 2\left(b+1\right)\left(a-b\right)^{2}}{2\left(b+1\right)\left(a-b\right)^{2}}$$

Since \frac{\left(a-b\right)\left(ab+a\right)\left(a^{2}-ab\right)}{2\left(b+1\right)\left(a-b\right)^{2}} and \frac{b\times 2\left(b+1\right)\left(a-b\right)^{2}}{2\left(b+1\right)\left(a-b\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(a-b\right)\left(ab+a\right)\left(a^{2}-ab\right)-b\times 2\left(b+1\right)\left(a-b\right)^{2}}{2\left(b+1\right)\left(a-b\right)^{2}}$$

Do the multiplications in \left(a-b\right)\left(ab+a\right)\left(a^{2}-ab\right)-b\times 2\left(b+1\right)\left(a-b\right)^{2}.

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

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

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

Factor the expressions that are not already factored in \frac{a^{4}b-2b^{3}+b^{3}a^{2}-2a^{3}b^{2}+a^{4}+4b^{3}a-2a^{3}b-b^{2}a^{2}-2b^{4}+4b^{2}a-2ba^{2}}{2\left(b+1\right)\left(a-b\right)^{2}}.

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

Cancel out \left(b+1\right)\left(a-b\right)^{2} in both numerator and denominator.

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