9) \( \frac{8 x^{3}+64 y^{3}}{x^{2}-x y-6 y^{2}} \)

To simplify the expression \( \frac{8 x^{3}+64 y^{3}}{x^{2}-x y-6 y^{2}} \), let's start with the numerator. We can factor \( 8 \) out of the numerator: \[ 8 (x^{3} + 8y^{3}) \] Next, notice that \( x^3 + 8y^3 \) can be factored using the sum of cubes: \[ x^3 + (2y)^3 = (x + 2y)(x^2 - 2xy + 4y^2) \] So, the numerator becomes: \[ 8 (x + 2y)(x^2 - 2xy + 4y^2) \] Now, let’s look at the denominator \( x^{2} - x y - 6 y^{2} \). We can factor this as a quadratic in terms of \( x \): This can be factored as: \[ (x - 3y)(x + 2y) \] Putting it all together, we have: \[ \frac{8 (x + 2y)(x^2 - 2xy + 4y^2)}{(x - 3y)(x + 2y)} \] Since \( (x + 2y) \) is in both the numerator and the denominator, we can cancel it out (as long as \( x + 2y \neq 0 \)): \[ \frac{8 (x^2 - 2xy + 4y^2)}{x - 3y} \] This is our simplified expression, provided \( x + 2y \neq 0 \): \[ \frac{8 (x^2 - 2xy + 4y^2)}{x - 3y} \] Remember to check for any restrictions on the values of \( x \) and \( y \) that could make any denominators zero!