How would you find the term x^2*y*z^5 in the expansion of (x-2y+z)^8? First you have to expand ((x-2y)+z)^8.
I understand how to find the coefficient but I don't really understand how to find the term.
Try to do it in two steps: as $\displaystyle (a+b)^n=\sum\limits_{k=0}^n\binom {n}{k}a^kb^{n-k}$, we get:
$\displaystyle ((x-2y)+z)^8=\sum\limits_{k=0}^n\binom {8}{k}(x-2y)^kz^{8-k}$
So you must choose k = 3 above to get $\displaystyle z^5$:$\displaystyle \binom {8}{3}(x-2y)^3z^5$ , and now you must choose the correct power in $\displaystyle (x-2y)^3$ to finally get $\displaystyle x^2yz^5$
Tonio