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:
Answer & explanation
Correct answer: option 3
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