Solve 2/x^2+1+4/x^4+1+8/x^8-1

$$\frac { 2 } { x ^ { 2 } + 1 } + \frac { 4 } { x ^ { 4 } + 1 } + \frac { 8 } { x ^ { 8 } - 1 }$$

Evaluate

$$\frac{2}{x^{2}-1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x^{2}+1 and x^{4}+1 is \left(x^{2}+1\right)\left(x^{4}+1\right). Multiply \frac{2}{x^{2}+1} times \frac{x^{4}+1}{x^{4}+1}. Multiply \frac{4}{x^{4}+1} times \frac{x^{2}+1}{x^{2}+1}.

$$\frac{2\left(x^{4}+1\right)}{\left(x^{2}+1\right)\left(x^{4}+1\right)}+\frac{4\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{4}+1\right)}+\frac{8}{x^{8}-1}$$

Since \frac{2\left(x^{4}+1\right)}{\left(x^{2}+1\right)\left(x^{4}+1\right)} and \frac{4\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{4}+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{2\left(x^{4}+1\right)+4\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{4}+1\right)}+\frac{8}{x^{8}-1}$$

Do the multiplications in 2\left(x^{4}+1\right)+4\left(x^{2}+1\right).

$$\frac{2x^{4}+2+4x^{2}+4}{\left(x^{2}+1\right)\left(x^{4}+1\right)}+\frac{8}{x^{8}-1}$$

Combine like terms in 2x^{4}+2+4x^{2}+4.

$$\frac{2x^{4}+6+4x^{2}}{\left(x^{2}+1\right)\left(x^{4}+1\right)}+\frac{8}{x^{8}-1}$$

Factor x^{8}-1.

$$\frac{2x^{4}+6+4x^{2}}{\left(x^{2}+1\right)\left(x^{4}+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x^{2}+1\right)\left(x^{4}+1\right) and \left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right) is \left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right). Multiply \frac{2x^{4}+6+4x^{2}}{\left(x^{2}+1\right)\left(x^{4}+1\right)} times \frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}.

$$\frac{\left(2x^{4}+6+4x^{2}\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)}$$

Since \frac{\left(2x^{4}+6+4x^{2}\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)} and \frac{8}{\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(2x^{4}+6+4x^{2}\right)\left(x-1\right)\left(x+1\right)+8}{\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)}$$

Do the multiplications in \left(2x^{4}+6+4x^{2}\right)\left(x-1\right)\left(x+1\right)+8.

$$\frac{2x^{6}-2x^{4}+6x^{2}-6+4x^{4}-4x^{2}+8}{\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)}$$

Combine like terms in 2x^{6}-2x^{4}+6x^{2}-6+4x^{4}-4x^{2}+8.

$$\frac{2x^{6}+2x^{4}+2x^{2}+2}{\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)}$$

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

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

Cancel out \left(x^{2}+1\right)\left(x^{4}+1\right) in both numerator and denominator.

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

Expand \left(x-1\right)\left(x+1\right).

$$\frac{2}{x^{2}-1}$$

Factor

$$\frac{2}{x^{2}-1}$$