Best way to expand $(2+x-x^2)^6$ 
I've completed part $(a)$ and gotten:
$64+192y+240y^2+160y^3+...$
Using intuition I substituted $x-x^2$ for $y$ and started listing the values for :
$y, y^2 $ and $y^3,$ in terms of $x$.
$y=(x-x^2)\\y^2=(x-x^2)^2 = x^2-2x^3+x^4;\\y^3 = (x-x^2)^3 = (x-x^2)(x^2-2x^3+x^4) = \;...$
Everything became complicated before I had even started to substitute these new-found values back into the polynomial devised from part $(a)$; for a 3 - out of 75 - mark question this seems extremely over-complicated. Am I doing it inefficiently / incorrectly; is there a pre-defined or easier method; or is it simple expansion and substitution that I'm too lazy to complete?
 A: You have everything you need.
Just multiply out $y^3$ as you have it and you'll get something with $x^3$ and higher powers of $x$.
Then, note that $y^4$ has terms of $x^4$ and higher, so they're all in the dot-dot-dot, which is Not Your Problem.
A: Here  is another proposal about how to calculate example 4b). We start with

Some considerations
  
  
*
  
*At first it was a good idea from OP to start with 4a.). Due to the similarity of the somewhat simpler expression
  \begin{align*}
(2+y)^6=64+192y+240y^2+160y^3+\cdots\tag{1}
\end{align*}
  compared with $(2+x-x^2)^6$ we can use the substitution
  $$y:=x-x^2$$
  and the result is
  \begin{align*}
(2+x-x^2)^2=64+192(x-x^2)+240(x-x^2)^2+160(x-x^2)^3+\cdots\tag{2}
\end{align*}
  
*Since we only need an expansion with powers up to $x^3$ we don't need any terms $(x-x^2)^n$ with $n>3$.
  
*We also recall the binomial formulas $(a+b)^n$ for $n=2,3$
  \begin{align*}
(a+b)^2&=a^2+2ab+b^2\\
(a+b)^3&=a^3+3a^2b+3ab^2+b^3
\end{align*}
  which we will use when expanding (2).

Now we have all the ingredients to effectively expand (2) up to powers of $x^3$.

The calculation
We obtain from (1) with the substitution: $y:= x-x^2$
  \begin{align*}
(2+x-x^2)^2&=64+192(x-x^2)+240(x-x^2)^2+160(x-x^2)^3+\cdots\\
&=64+192(x-x^2)+240(x^2-2x^3)+160(x^3)+\cdots\tag{3}\\
&=64+192x+(-192+240)x^2+(240\cdot(-2)+160)x^3+\cdots\tag{4}\\
&=64+192x+48x^2-320x^3+\cdots
\end{align*}
  and we're done here.

Comment:


*

*In (3) we expand the binomial formula up to third powers of $x$; everything else is put to $+\cdots$.

*In (4) we collect terms with equal powers.
A: I think I would say
$((x+2)(x-1))^6 = (x+2)^6(x-1)^6\\
(64 + 192x + 240x^2 + 160x^3 ...)(1 - 6x + 15x^2 - 20x^3 ...)$
I don't care about any powers bigger than 3
$64 + (64(-6) + 192)x + (64*15+192(-6) + 240)x^2 + (64(-20)+ 192*15 + 240*(-6) + 160)x^3...\\
64 - 192 x + 48 x^2 + 320x^3...$
