Cube of summation of independent (uncorrelated) random variables Basically, I am trying to prove an equality involving skewness of summation of random indepdent variables. However, I am stuck with the part after expressing sum of cube to $E[X^3] - 3E[X^2]\mu +2\mu^3$. I tried to solve using similar techniques for solving variances but now I am lost. And I could not find any helpful clue or proof. 
Given setting is as the following:
When $X = x_1 + x_2 + ... + x_n$, how to compute $E[X^3]$?
Also, how to express $E[X^3]$ using triple summations with $i, j, k = 1, 2, ... , n$ when $i, j, \& k$ may (or may not - possibly in both cases)  equivalent? 
Many thanks in advance. 


Also, how to express $E[X^3]$ using triple summations with $i, j, k = 1, 2, ... , n$ when $i, j, \& k$ may (or may not - possibly in both cases)  equivalent? 

$$\begin{align}\mathsf E(X^3) ~=~& \mathsf E({\sum}_{i,j,k} X_iX_jX_k) \\[1ex]~=~& {\sum}_i\mathsf E(X_i^3)+3{\sum}_{(i,j,k): i=j , j\neq k}\mathsf E(X_i^2X_k)+{\sum}_{(i,j,k):i\neq j,j\neq k,k\neq i}\mathsf E(X_iX_jX_k)\\[1ex]~=~& n\,\mathsf E(X_i^3)+3\,n\,(n-1)\,\mathsf E(X_i^2X_k)+n\,(n-1)\,(n-2)\,\mathsf E(X_iX_jX_k)\end{align}$$
Assuming these are independent and identically distributed random variables.
$$\begin{align}\mathsf E(X^3) ~=~& n\,\mathsf E(X_i^3)+3\,n\,(n-1)\,\mathsf E(X_i^2)\,\mathsf E(X_i)+n\,(n-1)\,(n-2)\,\mathsf E(X_i)^3\end{align}$$
