The given expression is:

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

Step 1: Factor the denominator

We can use the difference of cubes formula to factor the denominator. The difference of cubes formula is:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Here, $x^3 - y^3$ is a difference of cubes, so applying the formula gives:

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

Step 2: Simplify the expression

Now, substitute this factorization into the original expression:

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

Since $x - y$ appears in both the numerator and denominator, and assuming $x \neq y$ (to avoid division by zero), you can cancel $x - y$ from both sides:

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

Final result:

The simplified form of the expression is:

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

Would you like any more details or further explanation on this?

Here are 5 related questions to expand on this:

1. What happens if $x = y$?
2. How would the expression change if we had $x^3 + y^3$ instead of $x^3 - y^3$?
3. Can you explain the sum of cubes formula?
4. How do you factor more complex polynomials involving cubes?
5. How does the factorization of higher degree polynomials work?

Tip: When factoring polynomials, always look for common patterns like the difference of squares or cubes to simplify the expressions easily.