Binomial Series C4

(a+b+c+d)(e+f+g+h)=(ae+af+ag+ah+be+bf+bg+bh+ce+cf+cg+ch+de+df+dg+dh)

Stuck on exercise in textbook.

I understand how to expand the problem.

$\frac{(x+1)^3}{(2-x)}$

$(x+1)^3(2-x)^-1$

When expanded (to fourth term) i get $(1+3x + 3x^2 + x^3)(\frac{1}{2} + \frac {x}{4} + \frac{x^2}{8} + \frac{x^3}{16})$

Now i don't really understand how to combine the two expansions i've found to make one solution as the final answer.

The expansion works for only (1+x)^n expansion ..
so initially you had to take out a 1/2 and you end up with $(1-\frac{x}{2})^{-1}$ and expand..
and multiply each of them together untill get your fourth term..

I did that already.

Its combining the two expansions i was having trouble with.

what is the term they re looking for ?i.e are they looking for the coefficient of x^2,x^3,x^4??

They're looking for all the terms up to term 4. I've found them for both expansions and was finidng difficulty in putting them together to create one final solution.

would you mind posting it here ..

So multiply and add the like terms together?

It's in the first post.