Solve b^2+2/b^4-1+3/1-b^4

$$\frac { b ^ { 2 } + 2 } { b ^ { 4 } - 1 } + \frac { 3 } { 1 - b ^ { 4 } }$$

Evaluate

$$\frac{1}{b^{2}+1}$$

Short Solution Steps

Factor b^{4}-1. Factor 1-b^{4}.

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

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)\left(b^{2}+1\right) and \left(b-1\right)\left(b+1\right)\left(-b^{2}-1\right) is \left(b-1\right)\left(b+1\right)\left(b^{2}+1\right). Multiply \frac{3}{\left(b-1\right)\left(b+1\right)\left(-b^{2}-1\right)} times \frac{-1}{-1}.

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

Since \frac{b^{2}+2}{\left(b-1\right)\left(b+1\right)\left(b^{2}+1\right)} and \frac{3\left(-1\right)}{\left(b-1\right)\left(b+1\right)\left(b^{2}+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in b^{2}+2+3\left(-1\right).

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

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

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

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

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

Cancel out \left(b-1\right)\left(b+1\right) in both numerator and denominator.

$$\frac{1}{b^{2}+1}$$

Expand

$$\frac{1}{b^{2}+1}$$

Short Solution Steps

Factor b^{4}-1. Factor 1-b^{4}.

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

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)\left(b^{2}+1\right) and \left(b-1\right)\left(b+1\right)\left(-b^{2}-1\right) is \left(b-1\right)\left(b+1\right)\left(b^{2}+1\right). Multiply \frac{3}{\left(b-1\right)\left(b+1\right)\left(-b^{2}-1\right)} times \frac{-1}{-1}.

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

Since \frac{b^{2}+2}{\left(b-1\right)\left(b+1\right)\left(b^{2}+1\right)} and \frac{3\left(-1\right)}{\left(b-1\right)\left(b+1\right)\left(b^{2}+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in b^{2}+2+3\left(-1\right).

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

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

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

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

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

Cancel out \left(b-1\right)\left(b+1\right) in both numerator and denominator.

$$\frac{1}{b^{2}+1}$$