Question

Expand the first 3 (three) terms of the following: (2z^3-3y^3)^6

Answer 1

Please help, I don't understand how to do this

Answer 2

Recall the Binomial Theorem:
(a+b)^6 = a^6 + C(6,1)a^5 b + C(6,2)a^4 b^2 + ...
so plug in a=2z^3 and b = -3y^3 to get
(2z^3)^6 + 6(2z^3)^5 (-3y^3) + 15(2z^3)^4 (-3y^3)^2
64 z^18 - 576 z^15 y^3 + 2160 z^12 y^6