Solve x^3-2x(x+1/x^3-1

$$\frac { x ^ { 3 } - 2 x ( x + 1 } { x ^ { 3 } - 1 }$$

Evaluate

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

Short Solution Steps

Factor x^{3}-1.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{3} times \frac{\left(x-1\right)\left(x^{2}+x+1\right)}{\left(x-1\right)\left(x^{2}+x+1\right)}.

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

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

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

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

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

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

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

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

$$\frac{-2x+x^{6}-x^{3}-2x^{2}}{x^{3}-1}$$

Expand

$$\frac{x^{6}-x^{3}-2x^{2}-2x}{x^{3}-1}$$

Short Solution Steps

Factor x^{3}-1.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{3} times \frac{\left(x-1\right)\left(x^{2}+x+1\right)}{\left(x-1\right)\left(x^{2}+x+1\right)}.

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

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

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

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

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

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

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

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

$$\frac{-2x+x^{6}-x^{3}-2x^{2}}{x^{3}-1}$$