Answer :
From Binomial Theorem (x-2y)³:
([tex] (\frac{3}{0}) x^{3}+( \frac{3}{1} ) x^{2} *(-2y)+( \frac{3}{2})x*( -2y)^{2} +( \frac{3}{3}) x^{0} (-2y)^{3} =[/tex][tex] \frac{3!}{0!3!} x^{3}- \frac{3!}{1!2!} x^{2} 2y- \frac{3!}{2!1!}x4 y^{2}- \frac{3!}{3!0!}8 y^{3} [/tex]=x³-6x²y-12xy²-8y³
