Find the value of a, of b and of c in the identity $$x^{3}+3x ^{2}-2x+16-c(x+2) = ax ^{2}(x-1)+b(x-2)^{2}(x-1)$$

Question

anonymous · Answer

i don't see a snap way to do this
sometimes it is easy by picking a nice value for $x$ but i can't think of one here

anonymous · Answer

you can either multiply all this junk out and equate like coefficients
or you can replace $x$ by say 0, 1, 2 and get three equations in $a, b,c$

anonymous · Answer

I tried the method of comparing coefficients but I go an extremely weird answer and along the way had 6=0 :/

anonymous · Answer

put $x=1$ first

anonymous · Answer

you get 
$$1+3-2+16-3c=0$$

anonymous · Answer

oh right, okay

anonymous · Answer

i get 
$$c=6$$ and then you can find exactly what the left side is