Proof by Characteristic Function If $ X_1, X_2, \ldots,X_n$ are independent random variables variables with expectation $0$ and finite third moments.
Show that $$E((X_1+X_2+\cdots+X_n)^3) = EX_1^3+EX_2^3+\cdots+EX_n^3$$ using the aid of characteristic functions.$$$$
 A: One way to do this is by a simple induction on the number of terms after proving it works when $n=2$.
$$
E((X_1+X_2)^3) = E(X_1^3)+3E(X_1^2X_2)+3E(X_1X_2^2)+E(X_2^3).
$$
Because of independence this becomes
$$
E(X_1^3)+3E(X_1^2)E(X_2)+3E(X_1)E(X_2^2)+E(X_2^3).
$$
Then the middle two terms are $0$ because each has a factor that is $0$.
(But this doesn't fully answer the question because it doesn't use characteristic functions.  More later, maybe . . . . .)
Later edit: The characteristic function of random variable $X$ is
$$
\chi_X(t) = E\left( e^{itX} \right).
$$
It follows that
$$
\chi_X'''(t) = -iE\left(X^3 e^{itX} \right),
$$
so that $\chi_X'''(0) = -iE(X^3)$.
So we get
$$
\chi_{X_1+X_2}(t) = \chi_{X_1}(t)\chi_{X_2}(t),
$$
and then
$$
\chi_{X_1+X_2}'''(t) = \chi_{X_1}'''(t)\chi_{X_2}(t) + 3\chi_{X_1}''(t)\chi_{X_2}'(t) + 3\chi_{X_1}'(t)\chi_{X_2}''(t) + \chi_{X_1}(t)\chi_{X_2}'''(t).
$$
\begin{align}
\chi_{X_1+X_2}'''(0) & = \chi_{X_1}'''(0)\chi_{X_2}(0) + 3\chi_{X_1}''(t)\chi_{X_2}'(0) + 3\chi_{X_1}'(0)\chi_{X_2}''(0) + \chi_{X_1}(0)\chi_{X_2}'''(0) \\[6pt]
& = \chi_{X_1}'''(0)\cdot1 + 0 + 0 + 1\cdot\chi_{X_2}'''(0).
\end{align}
Then procede as above.
