Then $E(X_1+X_2+X_3+X_4)^4$ equals? Let $X_1, X_2, X_3, X_4$ are i.i.d random variable taking values $1$ and $-1$ with probability $1/2$ each. Then $E(X_1+X_2+X_3+X_4)^4$  equals?
I see that each $X_i$ is  standard normal and  so $X_1+X_2+X_3+X_4$ is a normal variable with mean $0$ and variance $4.$ I find the answer $48$ using MGF. But the answer is $76$. Where I gone wrong? thanks.

This seems one of those problems where just enumerating the possibilities is quick (even if dirty):
X1 X2  X3  X4  Sum Sum^4
-1 -1  -1  -1  -4  256
-1 -1  -1   1  -2   16
-1 -1   1  -1  -2   16
-1 -1   1   1   0    0
-1  1  -1  -1  -2   16
-1  1  -1   1   0    0
-1  1   1  -1   0    0
-1  1   1   1   2   16
1 -1  -1  -1  -2   16
1 -1  -1   1   0    0
1 -1   1  -1   0    0
1 -1   1   1   2   16
1  1  -1  -1   0    0
1  1  -1   1   2   16
1  1   1  -1   2   16
1  1   1   1   4  256
Now sum the last column and divide by $16$ as each row is equally probable to get an answer of $40$.
So either you haven't copied the question right or the answer your book has is wrong...
To double check, alternately, 
$$(X_1+X_2+X_3+X_4)^4 = \sum X_1^4 + 4\sum X_1^3X_2 + 6\sum X_1^2X_2^2+12\sum X_1X_2^2X_3 + 24X_1X_2X_3X_4$$
(where $\sum$ is used to denote sum of similar symmetric terms - however we dont need to count most of them, as $E(X_i^n) \in \{0, 1\}$ as $n$ is odd or even, and by independence we can multiply the expectations).
$$\implies E[(X_1+X_2+X_3+X_4)^4] = 4 + 0 + 6\times \binom42+0 + 0=40$$
