Best way to expand $(2+x-x^2)^6$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

$64 + (64(-6) + 192)x + (64*15+192(-6) + 240)x^2 + (64(-20)+ 192*15 + 240*(-6) + 160)x^3...\\ 64 - 192 x + 48 x^2 + 320x^3...$