$\frac{x^2(x-4)^2}{(x+4)^2-4 x} \div \frac{\left(x^2-4 x\right)^3}{(x+4)^2} \times \frac{64-x^3}{16-x^2}$ is equal to
Answer & explanation
Correct answer: option 3
A detailed explanation for this question is coming soon.
$\frac{x^2(x-4)^2}{(x+4)^2-4 x} \div \frac{\left(x^2-4 x\right)^3}{(x+4)^2} \times \frac{64-x^3}{16-x^2}$ is equal to
Correct answer: option 3
A detailed explanation for this question is coming soon.