On simplification, $\frac{x^3-y^3}{x\left[(x+y)^2-3 x y\right]} \div \frac{y\left[(x-y)^2+3 x y\right]}{x^3+y^3} \times \frac{(x+y)^2-(x-y)^2}{x^2-y^2}$ is equal to:

$\frac{1}{4}$ $\frac{1}{2}$ 4 1

4