Question

help me solving this using special products...
[x^2-(3x+2)]^3

Answer 1

Given (a + b) then (a + b)³ = a³ + 3a²b + 3ab² + b³.
\[(x ^{2})^{3}+3(x ^{2})^{2}(-3x-2)+3(x ^{2})(-3x-2)^{2}+(-3x-2)^2
x^6-9x^5+21x^4+9x^3-42x^2-36x-8

Answer 2

whoops. in the long middle setup, the last term should be cubed.