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\( \begin{array}{l} \text { Simplify the algebraic expression. }\\ \frac{x^{2} y+x y^{2}+y^{3}}{\frac{x^{2}}{y}-\frac{y^{2}}{x}}\\ \frac{x^{2} y+x y^{2}+y^{3}}{\frac{x^{2}}{y}-\frac{y^{2}}{x}}= \end{array} \)

Ask by Washington Klein. in the United States
Mar 08,2025

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The simplified expression is \( \frac{y^{2}x}{x-y} \).

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To simplify the given expression, we start with the numerator \(x^2y + xy^2 + y^3\) and the denominator \(\frac{x^2}{y} - \frac{y^2}{x}\). **Numerator:** We can factor out \(y\) from the first two terms: \[ x^2y + xy^2 + y^3 = y(x^2 + xy) + y^3 = y(x^2 + xy + y^2). \] **Denominator:** To combine the fractions in the denominator, we find a common denominator: \[ \frac{x^2}{y} - \frac{y^2}{x} = \frac{x^3 - y^3}{xy}, \] where \(x^3 - y^3\) can be factored using the difference of cubes: \[ x^3 - y^3 = (x - y)(x^2 + xy + y^2). \] Now plugging the numerator and denominator back into the expression gives us: \[ \frac{y(x^2 + xy + y^2)}{\frac{(x - y)(x^2 + xy + y^2)}{xy}} = \frac{y(x^2 + xy + y^2) \cdot xy}{(x - y)(x^2 + xy + y^2)}. \] Assuming \(x^2 + xy + y^2 \neq 0\) (which typically holds), we can cancel \(x^2 + xy + y^2\) from numerator and denominator: \[ = \frac{y \cdot xy}{x - y} = \frac{xy^2}{x - y}. \] Thus, the expression simplifies to: \[ \frac{xy^2}{x - y}. \]

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