Solve (3a-3ab+b-b^3/64a^2*8ab+8a/1-b^2):(9a+3b/4a(2b+6):2(b+1)-(b-1)/b^2+6b+9)

$$( \frac { 3 a - 3 a b + b - b ^ { 3 } } { 64 a ^ { 2 } } \cdot \frac { 8 a b + 8 a } { 1 - b ^ { 2 } } ) : ( \frac { 9 a + 3 b } { 4 a ( 2 b + 6 ) } : \frac { 2 ( b + 1 ) - ( b - 1 ) } { b ^ { 2 } + 6 b + 9 } )$$

Evaluate

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{8ab+8a}{1-b^{2}}.

$$\frac{\frac{3a-3ab+b-b^{3}}{64a^{2}}\times \frac{8a\left(b+1\right)}{\left(b-1\right)\left(-b-1\right)}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Extract the negative sign in 1+b.

$$\frac{\frac{3a-3ab+b-b^{3}}{64a^{2}}\times \frac{-8a\left(-b-1\right)}{\left(b-1\right)\left(-b-1\right)}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Cancel out -b-1 in both numerator and denominator.

$$\frac{\frac{3a-3ab+b-b^{3}}{64a^{2}}\times \frac{-8a}{b-1}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Multiply \frac{3a-3ab+b-b^{3}}{64a^{2}} times \frac{-8a}{b-1} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{\left(3a-3ab+b-b^{3}\right)\left(-8\right)a}{64a^{2}\left(b-1\right)}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Cancel out 8a in both numerator and denominator.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Divide \frac{9a+3b}{4a\left(2b+6\right)} by \frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9} by multiplying \frac{9a+3b}{4a\left(2b+6\right)} by the reciprocal of \frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{\left(9a+3b\right)\left(b^{2}+6b+9\right)}{4a\left(2b+6\right)\left(2\left(b+1\right)-\left(b-1\right)\right)}}$$

Factor the expressions that are not already factored in \frac{\left(9a+3b\right)\left(b^{2}+6b+9\right)}{4a\left(2b+6\right)\left(2\left(b+1\right)-\left(b-1\right)\right)}.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{3\left(3a+b\right)\left(b+3\right)^{2}}{2\times 4a\left(b+3\right)^{2}}}$$

Cancel out \left(b+3\right)^{2} in both numerator and denominator.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{3\left(3a+b\right)}{2\times 4a}}$$

Multiply 2 and 4 to get 8.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{3\left(3a+b\right)}{8a}}$$

Divide \frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)} by \frac{3\left(3a+b\right)}{8a} by multiplying \frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)} by the reciprocal of \frac{3\left(3a+b\right)}{8a}.

$$\frac{-\left(-3ab+3a-b^{3}+b\right)\times 8a}{8a\left(b-1\right)\times 3\left(3a+b\right)}$$

Cancel out 8a in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Extract the negative sign in 1-b.

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

Cancel out b-1 in both numerator and denominator.

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

Expand the expression.

$$\frac{3a+b^{2}+b}{9a+3b}$$

Expand

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{8ab+8a}{1-b^{2}}.

$$\frac{\frac{3a-3ab+b-b^{3}}{64a^{2}}\times \frac{8a\left(b+1\right)}{\left(b-1\right)\left(-b-1\right)}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Extract the negative sign in 1+b.

$$\frac{\frac{3a-3ab+b-b^{3}}{64a^{2}}\times \frac{-8a\left(-b-1\right)}{\left(b-1\right)\left(-b-1\right)}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Cancel out -b-1 in both numerator and denominator.

$$\frac{\frac{3a-3ab+b-b^{3}}{64a^{2}}\times \frac{-8a}{b-1}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Multiply \frac{3a-3ab+b-b^{3}}{64a^{2}} times \frac{-8a}{b-1} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{\left(3a-3ab+b-b^{3}\right)\left(-8\right)a}{64a^{2}\left(b-1\right)}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Cancel out 8a in both numerator and denominator.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{\frac{9a+3b}{4a\left(2b+6\right)}}{\frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}}}$$

Divide \frac{9a+3b}{4a\left(2b+6\right)} by \frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9} by multiplying \frac{9a+3b}{4a\left(2b+6\right)} by the reciprocal of \frac{2\left(b+1\right)-\left(b-1\right)}{b^{2}+6b+9}.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{\left(9a+3b\right)\left(b^{2}+6b+9\right)}{4a\left(2b+6\right)\left(2\left(b+1\right)-\left(b-1\right)\right)}}$$

Factor the expressions that are not already factored in \frac{\left(9a+3b\right)\left(b^{2}+6b+9\right)}{4a\left(2b+6\right)\left(2\left(b+1\right)-\left(b-1\right)\right)}.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{3\left(3a+b\right)\left(b+3\right)^{2}}{2\times 4a\left(b+3\right)^{2}}}$$

Cancel out \left(b+3\right)^{2} in both numerator and denominator.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{3\left(3a+b\right)}{2\times 4a}}$$

Multiply 2 and 4 to get 8.

$$\frac{\frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)}}{\frac{3\left(3a+b\right)}{8a}}$$

Divide \frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)} by \frac{3\left(3a+b\right)}{8a} by multiplying \frac{-\left(-3ab+3a-b^{3}+b\right)}{8a\left(b-1\right)} by the reciprocal of \frac{3\left(3a+b\right)}{8a}.

$$\frac{-\left(-3ab+3a-b^{3}+b\right)\times 8a}{8a\left(b-1\right)\times 3\left(3a+b\right)}$$

Cancel out 8a in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Extract the negative sign in 1-b.

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

Cancel out b-1 in both numerator and denominator.

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

Expand the expression.

$$\frac{3a+b^{2}+b}{9a+3b}$$