Sum of Coefficients in a Polynomial Find the sum of the coefficients of the terms in the expansion of $(2x+3y-3z)^7$.
I know how to do this for binomials, but I was not able to apply the same logic to a trinomial. 
I believe my other method is faulty, but here it is: $2+3-3 = 2$, and $2^7 = 128$, which is the sum of the coefficients.
Please tell me if there is a basic method that can be used to solve this problem in a general case and if my method is correct/faulty.
 A: Set $$x = y = z = 1$$ With these values the expansion is the sum of the coefficient.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Find the sum of the coefficients of the terms in the expansion of
  $\ds{\pars{2x + 3y - 3z}^{7}:\ {\large ?}}$

\begin{align}&\color{#66f}{\large%
\sum_{a\ =\ 0}^{\infty}\sum_{b\ =\ 0}^{\infty}\sum_{c\ =\ 0}^{\infty}
{7! \over a!\, b!\, c!}2^{a}3^{b}\pars{-3}^{c}\delta_{a + b + c,7}}
\\[5mm]&=7!\sum_{a\ =\ 0}^{\infty}\sum_{b\ =\ 0}^{\infty}\sum_{c\ =\ 0}^{\infty}
{2^{a} \over a!}\,{3^{b} \over b!}\,{\pars{-3}^{c} \over c!}\oint_{\verts{z}\ =\ 1}
{1 \over z^{-a - b - c + 8}}\,{\dd z \over 2\pi\ic}
\\[5mm]&=7!\oint_{\verts{z}\ =\ 1}
{1 \over z^{8}}\sum_{a\ =\ 0}^{\infty}{\pars{2z}^{a} \over a!}
\sum_{b\ =\ 0}^{\infty}{\pars{3z}^{b} \over b!}\sum_{c\ =\ 0}^{\infty}{\pars{-3z}^{c} \over c!}\,{\dd z \over 2\pi\ic}
\\[5mm]&=7!\oint_{\verts{z}\ =\ 1}
{1 \over z^{8}}\,\expo{2z}\expo{3z}\expo{-3z}\,{\dd z \over 2\pi\ic}
=7!\oint_{\verts{z}\ =\ 1}{\expo{2z} \over z^{8}}\,{\dd z \over 2\pi\ic}
=7!\sum_{n\ =\ 0}^{\infty}{2^{n} \over n!}\
\overbrace{\oint_{\verts{z}\ =\ 1}{1 \over z^{8 - n}}\,{\dd z \over 2\pi\ic}}
^{\dsc{\delta_{n,7}}}
\\[5mm]&=7!\,{2^{7} \over 7!}=2^{7}=\color{#66f}{\Large 128}
\end{align}
A: Expanding (pun not intended) on WLOG's answer:
The expression $(2x+3y-3z)^7$ expands to
$$(2x+3y-3z)^7 = \sum_{\substack{0 \le i,j,k\\i+j+k=7}} (2x)^i(3y)^j(-3z)^k = \sum_{\substack{0\le i,j,k\\i+j+k=7}}2^i3^j(-3)^kx^iy^jz^k \tag{1}$$
(The notation indicates that we are summing over all combinations of nonnegative integers $i,j,k$ which sum to $7$).  Setting $x=y=z=1$ gives us
$$\sum_{\substack{0\le i,j,k\\i+j+k=7}}2^i3^j(-3)^k$$
which is exactly the sum of the coefficients.  But, on the other hand, by (1),
$$\sum_{\substack{0\le i,j,k\\i+j+k=7}}2^i3^j(-3)^k = (2+3-3)^7 = 128$$
as you observed.  Thus, the sum of the coefficients of the expansion is $128$.  Notice that this method could be extended to an arbitrary number of variables and choice of coefficients:
$$(a_1x_1 + \cdots + a_nx_n)^k = (a_1 + \cdots + a_n)^k$$
