Solve 1-(x-1/1-x)^2divx^2-x+1/x^2-2x+1

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

Evaluate

$$x\left(1-x\right)$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{1-x}{1-x}.

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

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

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

Do the multiplications in x\left(1-x\right)-1.

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

To raise \frac{x-x^{2}-1}{1-x} to a power, raise both numerator and denominator to the power and then divide.

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

Divide \frac{\left(x-x^{2}-1\right)^{2}}{\left(1-x\right)^{2}} by \frac{x^{2}-x+1}{x^{2}-2x+1} by multiplying \frac{\left(x-x^{2}-1\right)^{2}}{\left(1-x\right)^{2}} by the reciprocal of \frac{x^{2}-x+1}{x^{2}-2x+1}.

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

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

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

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

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

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

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

Combine like terms in x^{2}-x+1-2x^{3}+2x^{2}-2x+x^{4}-x^{3}+x^{2}-x^{4}+2x^{3}-x^{2}+x^{5}-2x^{4}+x^{3}+x^{3}-2x^{2}+x+x^{5}-2x^{4}+x^{3}-x^{6}+2x^{5}-x^{4}-x^{4}+2x^{3}-x^{2}+x^{3}-2x^{2}+x-x^{4}+2x^{3}-x^{2}-x^{2}+2x-1.

$$\frac{-4x^{2}+x+7x^{3}-7x^{4}+4x^{5}-x^{6}}{\left(1-x\right)^{2}\left(x^{2}-x+1\right)}$$

Factor the expressions that are not already factored in \frac{-4x^{2}+x+7x^{3}-7x^{4}+4x^{5}-x^{6}}{\left(1-x\right)^{2}\left(x^{2}-x+1\right)}.

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

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

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

Cancel out x^{2}-x+1 in both numerator and denominator.

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

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

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

Factor the expressions that are not already factored.

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

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

$$-x\left(x-1\right)$$

Expand the expression.

$$-x^{2}+x$$

Expand

$$x-x^{2}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{1-x}{1-x}.

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

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

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

Do the multiplications in x\left(1-x\right)-1.

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

To raise \frac{x-x^{2}-1}{1-x} to a power, raise both numerator and denominator to the power and then divide.

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

Divide \frac{\left(x-x^{2}-1\right)^{2}}{\left(1-x\right)^{2}} by \frac{x^{2}-x+1}{x^{2}-2x+1} by multiplying \frac{\left(x-x^{2}-1\right)^{2}}{\left(1-x\right)^{2}} by the reciprocal of \frac{x^{2}-x+1}{x^{2}-2x+1}.

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

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

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

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

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

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

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

Combine like terms in x^{2}-x+1-2x^{3}+2x^{2}-2x+x^{4}-x^{3}+x^{2}-x^{4}+2x^{3}-x^{2}+x^{5}-2x^{4}+x^{3}+x^{3}-2x^{2}+x+x^{5}-2x^{4}+x^{3}-x^{6}+2x^{5}-x^{4}-x^{4}+2x^{3}-x^{2}+x^{3}-2x^{2}+x-x^{4}+2x^{3}-x^{2}-x^{2}+2x-1.

$$\frac{-4x^{2}+x+7x^{3}-7x^{4}+4x^{5}-x^{6}}{\left(1-x\right)^{2}\left(x^{2}-x+1\right)}$$

Factor the expressions that are not already factored in \frac{-4x^{2}+x+7x^{3}-7x^{4}+4x^{5}-x^{6}}{\left(1-x\right)^{2}\left(x^{2}-x+1\right)}.

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

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

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

Cancel out x^{2}-x+1 in both numerator and denominator.

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

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

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

Factor the expressions that are not already factored.

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

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

$$-x\left(x-1\right)$$

Expand the expression.

$$-x^{2}+x$$