Let's solve the given problem.

The expression is:

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

Step 1: Simplify the first term

The first part is:

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

We know that $x^3 + y^3$ can be factored as:

$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$

Thus, the first term simplifies to:

\frac{(x + y)(x^2 - xy + y^2)}{x^2 - xy + y^2} = x + y \quad \text{(since x^2 - xy + y^2 cancels out)}

Step 2: Simplify the second term

The second term is:

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

We recognize that $x^2 - y^2$ is a difference of squares, which can be factored as:

$x^2 - y^2 = (x - y)(x + y)$

Thus, the second term simplifies to:

\frac{(x - y)(x + y)}{x + y} = x - y \quad \text{(since x + y cancels out)}

Step 3: Combine the two simplified terms

Now, we subtract the two simplified terms:

$(x + y) - (x - y) = x + y - x + y = 2y$

Final Answer:

The simplified result is $2y$, which corresponds to option B.

Would you like further details or have any other questions?

Here are 5 related questions to deepen your understanding:

1. How do you factor expressions like $x^3 + y^3$?
2. Why does factoring help in simplifying rational expressions?
3. Can you apply the same technique to simplify other polynomials?
4. How does the difference of squares work for higher powers?
5. What happens if $x = y$ in this equation?

Tip: Always check for common factoring patterns like difference of squares or sum/difference of cubes when simplifying algebraic expressions.