Solve frac{2+2c^{4}}{1-c^{4}}+4c^2/1+c^4+8c^6/1-c^8

$$\frac { 2 + 2 c ^ { 4 } } { 1 - c ^ { 4 } } + \frac { 4 c ^ { 2 } } { 1 + c ^ { 4 } } + \frac { 8 c ^ { 6 } } { 1 - c ^ { 8 } }$$

Evaluate

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

Short Solution Steps

Factor 1-c^{4}.

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

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

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

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

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

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

$$\frac{2c^{4}+2+2c^{8}+2c^{4}-4c^{6}-4c^{4}-4c^{5}-4c^{3}+4c^{5}+4c^{3}+4c^{4}+4c^{2}}{\left(c-1\right)\left(c+1\right)\left(-c^{2}-1\right)\left(c^{4}+1\right)}+\frac{8c^{6}}{1-c^{8}}$$

Combine like terms in 2c^{4}+2+2c^{8}+2c^{4}-4c^{6}-4c^{4}-4c^{5}-4c^{3}+4c^{5}+4c^{3}+4c^{4}+4c^{2}.

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

Factor 1-c^{8}.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(c-1\right)\left(c+1\right)\left(-c^{2}-1\right)\left(c^{4}+1\right) and \left(c-1\right)\left(c+1\right)\left(c^{2}+1\right)\left(-c^{4}-1\right) is \left(c-1\right)\left(c+1\right)\left(c^{2}+1\right)\left(c^{4}+1\right). Multiply \frac{4c^{4}+2c^{8}+2-4c^{6}+4c^{2}}{\left(c-1\right)\left(c+1\right)\left(-c^{2}-1\right)\left(c^{4}+1\right)} times \frac{-1}{-1}. Multiply \frac{8c^{6}}{\left(c-1\right)\left(c+1\right)\left(c^{2}+1\right)\left(-c^{4}-1\right)} times \frac{-1}{-1}.

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

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

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

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

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

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

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

Factor the expressions that are not already factored in \frac{-4c^{4}-2-2c^{8}-4c^{6}-4c^{2}}{\left(c-1\right)\left(c+1\right)\left(c^{2}+1\right)\left(c^{4}+1\right)}.

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

Extract the negative sign in -1-c^{2}.

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

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

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

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

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

Use the distributive property to multiply -2 by c^{2}+1.

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

Expand

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

Short Solution Steps

Factor 1-c^{4}.

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

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

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

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

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

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

$$\frac{2c^{4}+2+2c^{8}+2c^{4}-4c^{6}-4c^{4}-4c^{5}-4c^{3}+4c^{5}+4c^{3}+4c^{4}+4c^{2}}{\left(c-1\right)\left(c+1\right)\left(-c^{2}-1\right)\left(c^{4}+1\right)}+\frac{8c^{6}}{1-c^{8}}$$

Combine like terms in 2c^{4}+2+2c^{8}+2c^{4}-4c^{6}-4c^{4}-4c^{5}-4c^{3}+4c^{5}+4c^{3}+4c^{4}+4c^{2}.

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

Factor 1-c^{8}.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(c-1\right)\left(c+1\right)\left(-c^{2}-1\right)\left(c^{4}+1\right) and \left(c-1\right)\left(c+1\right)\left(c^{2}+1\right)\left(-c^{4}-1\right) is \left(c-1\right)\left(c+1\right)\left(c^{2}+1\right)\left(c^{4}+1\right). Multiply \frac{4c^{4}+2c^{8}+2-4c^{6}+4c^{2}}{\left(c-1\right)\left(c+1\right)\left(-c^{2}-1\right)\left(c^{4}+1\right)} times \frac{-1}{-1}. Multiply \frac{8c^{6}}{\left(c-1\right)\left(c+1\right)\left(c^{2}+1\right)\left(-c^{4}-1\right)} times \frac{-1}{-1}.

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

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

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

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

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

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

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

Factor the expressions that are not already factored in \frac{-4c^{4}-2-2c^{8}-4c^{6}-4c^{2}}{\left(c-1\right)\left(c+1\right)\left(c^{2}+1\right)\left(c^{4}+1\right)}.

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

Extract the negative sign in -1-c^{2}.

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

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

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

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

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

Use the distributive property to multiply -2 by c^{2}+1.

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