Solve frac{14b^{2}-5b}{2b+1}divb^2+b-2/b^3+8*b-1/b^3-2b^2

$$\frac { 14 b ^ { 2 } - 5 b } { 2 b + 1 } \div \frac { b ^ { 2 } + b - 2 } { b ^ { 3 } + 8 } \cdot \frac { b - 1 } { b ^ { 3 } - 2 b ^ { 2 } }$$

Evaluate

$$\frac{\left(14b-5\right)\left(b^{2}-2b+4\right)}{b\left(b-2\right)\left(2b+1\right)}$$

Short Solution Steps

Divide \frac{14b^{2}-5b}{2b+1} by \frac{b^{2}+b-2}{b^{3}+8} by multiplying \frac{14b^{2}-5b}{2b+1} by the reciprocal of \frac{b^{2}+b-2}{b^{3}+8}.

$$\frac{\left(14b^{2}-5b\right)\left(b^{3}+8\right)}{\left(2b+1\right)\left(b^{2}+b-2\right)}\times \frac{b-1}{b^{3}-2b^{2}}$$

Factor the expressions that are not already factored in \frac{\left(14b^{2}-5b\right)\left(b^{3}+8\right)}{\left(2b+1\right)\left(b^{2}+b-2\right)}.

$$\frac{b\left(14b-5\right)\left(b+2\right)\left(b^{2}-2b+4\right)}{\left(b-1\right)\left(b+2\right)\left(2b+1\right)}\times \frac{b-1}{b^{3}-2b^{2}}$$

Cancel out b+2 in both numerator and denominator.

$$\frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)}{\left(b-1\right)\left(2b+1\right)}\times \frac{b-1}{b^{3}-2b^{2}}$$

Multiply \frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)}{\left(b-1\right)\left(2b+1\right)} times \frac{b-1}{b^{3}-2b^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)\left(b-1\right)}{\left(b-1\right)\left(2b+1\right)\left(b^{3}-2b^{2}\right)}$$

Cancel out b-1 in both numerator and denominator.

$$\frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)}{\left(2b+1\right)\left(b^{3}-2b^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)}{\left(b-2\right)\left(2b+1\right)b^{2}}$$

Cancel out b in both numerator and denominator.

$$\frac{\left(14b-5\right)\left(b^{2}-2b+4\right)}{b\left(b-2\right)\left(2b+1\right)}$$

Expand the expression.

$$\frac{14b^{3}-33b^{2}+66b-20}{2b^{3}-3b^{2}-2b}$$

Expand

$$\frac{14b^{3}-33b^{2}+66b-20}{b\left(b-2\right)\left(2b+1\right)}$$

Short Solution Steps

Divide \frac{14b^{2}-5b}{2b+1} by \frac{b^{2}+b-2}{b^{3}+8} by multiplying \frac{14b^{2}-5b}{2b+1} by the reciprocal of \frac{b^{2}+b-2}{b^{3}+8}.

$$\frac{\left(14b^{2}-5b\right)\left(b^{3}+8\right)}{\left(2b+1\right)\left(b^{2}+b-2\right)}\times \frac{b-1}{b^{3}-2b^{2}}$$

Factor the expressions that are not already factored in \frac{\left(14b^{2}-5b\right)\left(b^{3}+8\right)}{\left(2b+1\right)\left(b^{2}+b-2\right)}.

$$\frac{b\left(14b-5\right)\left(b+2\right)\left(b^{2}-2b+4\right)}{\left(b-1\right)\left(b+2\right)\left(2b+1\right)}\times \frac{b-1}{b^{3}-2b^{2}}$$

Cancel out b+2 in both numerator and denominator.

$$\frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)}{\left(b-1\right)\left(2b+1\right)}\times \frac{b-1}{b^{3}-2b^{2}}$$

Multiply \frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)}{\left(b-1\right)\left(2b+1\right)} times \frac{b-1}{b^{3}-2b^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)\left(b-1\right)}{\left(b-1\right)\left(2b+1\right)\left(b^{3}-2b^{2}\right)}$$

Cancel out b-1 in both numerator and denominator.

$$\frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)}{\left(2b+1\right)\left(b^{3}-2b^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{b\left(14b-5\right)\left(b^{2}-2b+4\right)}{\left(b-2\right)\left(2b+1\right)b^{2}}$$

Cancel out b in both numerator and denominator.

$$\frac{\left(14b-5\right)\left(b^{2}-2b+4\right)}{b\left(b-2\right)\left(2b+1\right)}$$

Expand the expression.

$$\frac{14b^{3}-33b^{2}+66b-20}{2b^{3}-3b^{2}-2b}$$