Simplify the algebraic expression. $$ \frac{x^{3}+x^{2} y+x y^{2}}{\frac{x^{2}}{y}-\frac{y^{2}}{x}} $$ $$ \frac{x^{3}+x^{2} y+x y^{2}}{x^{2}} \frac{y^{2}}{y}-\frac{1}{x} $$

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We start with
$$
\frac{x^3+x^2y+xy^2}{\frac{x^2}{y}-\frac{y^2}{x}}.
$$

**Step 1. Factor the numerator.**

Factor out the common factor $$x$$:
$$
x^3+x^2y+xy^2 = x\,(x^2+xy+y^2).
$$

**Step 2. Simplify the denominator.**

Write the denominator with a common denominator. Notice that
$$
\frac{x^2}{y}-\frac{y^2}{x} = \frac{x^3}{xy}-\frac{y^3}{xy} = \frac{x^3-y^3}{xy}.
$$

**Step 3. Write the full expression.**

Now our expression becomes
$$
\frac{x(x^2+xy+y^2)}{\frac{x^3-y^3}{xy}}.
$$
Dividing by a fraction is equivalent to multiplying by its reciprocal:
$$
= x(x^2+xy+y^2) \cdot \frac{xy}{x^3-y^3}.
$$
Multiply the $$x$$ by the $$xy$$:
$$
= \frac{x^2y\,(x^2+xy+y^2)}{x^3-y^3}.
$$

**Step 4. Factor the difference of cubes in the denominator.**

Recall that
$$
x^3-y^3 = (x-y)(x^2+xy+y^2).
$$
Substitute this into our expression:
$$
= \frac{x^2y\,(x^2+xy+y^2)}{(x-y)(x^2+xy+y^2)}.
$$

**Step 5. Cancel the common factor.**

Cancel the common factor $$(x^2+xy+y^2)$$ (assuming it is nonzero):
$$
= \frac{x^2y}{x-y}.
$$

Thus, the simplified expression is
$$
\frac{x^2y}{x-y}.
$$
The simplified expression is $$ \frac{x^2 y}{x - y} $$.

Simplify the algebraic expression.

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