Solve xy^-2-yx^-2/y^-1-x^-1

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

Evaluate

$$\frac{x^{2}+xy+y^{2}}{xy}$$

Short Solution Steps

Factor the expressions that are not already factored.

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

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

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

Expand the expression.

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

Express \frac{1}{x}y^{2} as a single fraction.

$$\frac{x-x^{-2}y^{3}}{-\frac{y^{2}}{x}+y}$$

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

$$\frac{x-x^{-2}y^{3}}{-\frac{y^{2}}{x}+\frac{yx}{x}}$$

Since -\frac{y^{2}}{x} and \frac{yx}{x} have the same denominator, add them by adding their numerators.

$$\frac{x-x^{-2}y^{3}}{\frac{-y^{2}+yx}{x}}$$

Divide x-x^{-2}y^{3} by \frac{-y^{2}+yx}{x} by multiplying x-x^{-2}y^{3} by the reciprocal of \frac{-y^{2}+yx}{x}.

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

Factor the expressions that are not already factored.

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

Cancel out x-y in both numerator and denominator.

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

Expand the expression.

$$\frac{x+y+\frac{1}{x}y^{2}}{y}$$

Express \frac{1}{x}y^{2} as a single fraction.

$$\frac{x+y+\frac{y^{2}}{x}}{y}$$

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

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

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

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

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

$$\frac{\frac{x^{2}+xy+y^{2}}{x}}{y}$$

Express \frac{\frac{x^{2}+xy+y^{2}}{x}}{y} as a single fraction.

$$\frac{x^{2}+xy+y^{2}}{xy}$$

Expand

$$\frac{x^{2}+xy+y^{2}}{xy}$$

Short Solution Steps

Factor the expressions that are not already factored.

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

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

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

Expand the expression.

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

Express \frac{1}{x}y^{2} as a single fraction.

$$\frac{x-x^{-2}y^{3}}{-\frac{y^{2}}{x}+y}$$

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

$$\frac{x-x^{-2}y^{3}}{-\frac{y^{2}}{x}+\frac{yx}{x}}$$

Since -\frac{y^{2}}{x} and \frac{yx}{x} have the same denominator, add them by adding their numerators.

$$\frac{x-x^{-2}y^{3}}{\frac{-y^{2}+yx}{x}}$$

Divide x-x^{-2}y^{3} by \frac{-y^{2}+yx}{x} by multiplying x-x^{-2}y^{3} by the reciprocal of \frac{-y^{2}+yx}{x}.

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

Factor the expressions that are not already factored.

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

Cancel out x-y in both numerator and denominator.

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

Expand the expression.

$$\frac{x+y+\frac{1}{x}y^{2}}{y}$$

Express \frac{1}{x}y^{2} as a single fraction.

$$\frac{x+y+\frac{y^{2}}{x}}{y}$$

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

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

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

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

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

$$\frac{\frac{x^{2}+xy+y^{2}}{x}}{y}$$

Express \frac{\frac{x^{2}+xy+y^{2}}{x}}{y} as a single fraction.

$$\frac{x^{2}+xy+y^{2}}{xy}$$