Characteristic functions, show that $\mathrm{E}[X_1 + X_2 +...+X_n]^3 = \mathrm{E}[X_1^3] +...+\mathrm{E}[X_n^3]$ when $\mathrm{E}[X] = 0$ 
I want to show that for independent variables X_1, X_2,...,X_n, with expectation 0 and finite third moments,
$$\mathrm{E}[X_1 + X_2 +...+X_n]^3 = \mathrm{E}[X_1^3] + \mathrm{E}[X_2^3] +...+\mathrm{E}[X_n^3].$$

I have tried to show this, and it isn't working for me. I might even be completely off, since this is all pretty new to me. I am supposed to use characteristic functions to solve the problem.
I put $S_n = X_1 + X_2 + ... + X_n$, and I know that in general when the $k$:th moment exists:
$\mathrm{E}[Y^k] = i^{-k} \varphi_Y^{\left(k\right)}(0),$
so $\mathrm{E}[S_n] = \frac{1}{i} \varphi_{S_n}(0).$
I also know that $\varphi_{S_n}(0) = \prod_{k=1}^n \varphi_{X_k}(0),$
so I should have that $\mathrm{E}[S_n] = \frac{1}{i} \varphi_{X_1}(0) \dots \varphi_{X_n}(0) \Rightarrow (\mathrm{E}(S_n))^3 = \left(\frac{1}{i} \varphi_{X_1}(0) \dots \varphi_{X_n}(0)\right)^3.$
Furthermore, I have that
$\varphi_X(t) = 1 - \frac{1}{2}t^2 \sigma^2 + o(t^2)$ for $t \rightarrow 0$ when $\mathrm{E}[X] = 0,$ which I think means that $\varphi_X(0) = 1,$ although I'm not sure since it's not quite clear to me what the term $o(t^2)$ stands for.
So I figure I should have that $\mathrm{E}[S_n] = \frac{1}{i^3} (1).$
With $\mathrm{E}[X_1^3] + ... + \mathrm{E}[X_n^3] = i^{-3} \varphi_{X_1}^{\left(3\right)}(0) + ... + i^{-3} \varphi_{X_n}^{\left(3\right)}(0),$
and again using that $\varphi_X(0) = 1$ when $\mathrm{E}[X] = 0,$
that would lead me to
$\mathrm{E}[X_1^3] +...+\mathrm{E}[X_n] = n\frac{1}{i^3},$ which isn't the result I wanted to reach...
 A: Since the random variables are independent, we have that
$$
\operatorname E[X_iX_jX_k]=0
$$
for $i,j,k\in\{1,\ldots,n\}$ unless $i=j=k$. When we raise
$$
X_1+\ldots+X_n
$$
to the power of three, we obtain a sum of terms $cX_iX_jX_k$, where $i,j,k\in\{1,\ldots,n\}$ and $c$ is some arbitrary constant. When we take the expectations, all the terms cancel out except the terms $\operatorname EX_1^3,\ldots,\operatorname EX_n^3$. Hence, we have that
$$
\operatorname E[X_1+\ldots+X_n]^3=\operatorname EX_1^3+\ldots+\operatorname EX_n^3.
$$
A: Do you know the multinomial theorem? How about the binomial theorem?
Expanding $$(x_1 + ... + x_n)^3$$
gives us terms that look like
$$x_i^3, x_ix_jx_k, x_i^2x_k, x_ix_k^2$$
We use linearity to split the expectation of the cube of the sum into sums of expectations that look like
$$E[x_i^3], E[x_ix_jx_k], E[x_i^2x_k],E[x_ix_k^2]$$
We us independence to split the last three types of expectations into products of expectations:
$$E[x_i]E[x_j]E[x_k], E[x_i^2]E[x_k], $E[x_i]E[x_k^2]$$
The above are all zero.
