Solve frac{x^{2}-x+1}{x^{2}+x+1}+frac{2x(x-1)^{2}}{x^{4}+x^{2}+1}+frac{2x^{2}(x^2-1)^2}{x^8+x^4+1}

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

Evaluate

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

Short Solution Steps

Factor x^{4}+x^{2}+1.

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

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

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

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

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

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

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

Combine like terms in x^{4}-x^{3}+x^{2}-x^{3}+x^{2}-x+x^{2}-x+1+2x^{3}-4x^{2}+2x.

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

Factor x^{8}+x^{4}+1.

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

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

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

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

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

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

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

Combine like terms in x^{8}-x^{6}+x^{4}-x^{6}+x^{4}-x^{2}+x^{4}-x^{2}+1+2x^{6}-4x^{4}+2x^{2}.

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

Expand \left(x^{2}+x+1\right)\left(x^{2}-x+1\right)\left(x^{4}-x^{2}+1\right).

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

Expand

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

Short Solution Steps

Factor x^{4}+x^{2}+1.

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

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

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

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

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

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

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

Combine like terms in x^{4}-x^{3}+x^{2}-x^{3}+x^{2}-x+x^{2}-x+1+2x^{3}-4x^{2}+2x.

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

Factor x^{8}+x^{4}+1.

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

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

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

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

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

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

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

Combine like terms in x^{8}-x^{6}+x^{4}-x^{6}+x^{4}-x^{2}+x^{4}-x^{2}+1+2x^{6}-4x^{4}+2x^{2}.

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

Expand \left(x^{2}+x+1\right)\left(x^{2}-x+1\right)\left(x^{4}-x^{2}+1\right).

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