Solve (1-b/1+b-1+b/1-b+1+4b/1-b^2)*(b^2+2b+1)

$$( \frac { 1 - b } { 1 + b } - \frac { 1 + b } { 1 - b } + \frac { 1 + 4 b } { 1 - b ^ { 2 } } ) \cdot ( b ^ { 2 } + 2 b + 1 )$$

Evaluate

$$\frac{b+1}{1-b}$$

Short Solution Steps

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

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

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

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

Do the multiplications in \left(1-b\right)\left(-b+1\right)-\left(1+b\right)\left(b+1\right).

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

Combine like terms in 1-b+b^{2}-b-b-1-b^{2}-b.

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

Factor 1-b^{2}.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(b+1\right)\left(-b+1\right) and \left(b-1\right)\left(-b-1\right) is \left(b-1\right)\left(b+1\right). Multiply \frac{-4b}{\left(b+1\right)\left(-b+1\right)} times \frac{-1}{-1}. Multiply \frac{1+4b}{\left(b-1\right)\left(-b-1\right)} times \frac{-1}{-1}.

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

Since \frac{-\left(-4\right)b}{\left(b-1\right)\left(b+1\right)} and \frac{-\left(1+4b\right)}{\left(b-1\right)\left(b+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in -\left(-4\right)b-\left(1+4b\right).

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

Combine like terms in 4b-1-4b.

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

Express \frac{-1}{\left(b-1\right)\left(b+1\right)}\left(b^{2}+2b+1\right) as a single fraction.

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

Factor the expressions that are not already factored.

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

Cancel out b+1 in both numerator and denominator.

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

Expand the expression.

$$\frac{-b-1}{b-1}$$

Expand

$$\frac{b+1}{1-b}$$

Short Solution Steps

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

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

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

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

Do the multiplications in \left(1-b\right)\left(-b+1\right)-\left(1+b\right)\left(b+1\right).

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

Combine like terms in 1-b+b^{2}-b-b-1-b^{2}-b.

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

Factor 1-b^{2}.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(b+1\right)\left(-b+1\right) and \left(b-1\right)\left(-b-1\right) is \left(b-1\right)\left(b+1\right). Multiply \frac{-4b}{\left(b+1\right)\left(-b+1\right)} times \frac{-1}{-1}. Multiply \frac{1+4b}{\left(b-1\right)\left(-b-1\right)} times \frac{-1}{-1}.

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

Since \frac{-\left(-4\right)b}{\left(b-1\right)\left(b+1\right)} and \frac{-\left(1+4b\right)}{\left(b-1\right)\left(b+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in -\left(-4\right)b-\left(1+4b\right).

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

Combine like terms in 4b-1-4b.

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

Express \frac{-1}{\left(b-1\right)\left(b+1\right)}\left(b^{2}+2b+1\right) as a single fraction.

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

Factor the expressions that are not already factored.

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

Cancel out b+1 in both numerator and denominator.

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

Expand the expression.

$$\frac{-b-1}{b-1}$$