What is the mean and standard deviation of $5x^3 + 8$ when $X$ is the sum of numbers of two fair dice 
Consider the following experiment: Throw two fair dice sequentially and define a random variable $X$ as the sum of the numbers of the two dice.  Please compute the mean and standard deviation of $5X^3+8.$

I am able to come up with $E(X)$ being $7$, $\operatorname{Var} (X)$ being $5.83333$. But, I'm not sure if I can simply substitute $E(X) = 7$ into $5X^3 + 8$ to get the mean of $5X^3 + 8$. The same question for the s.d.
 A: Make a table and apply the definitions.
The expectation is found by multiplying each possible value by its chance and adding up those results.  The residuals are the differences between each possible value and the expectation. The variance is the expectation of the squared residuals.  The standard deviation is the square root of the variance.
To illustrate, imagine a die with three equally likely sides bearing the numbers $1,$ $2,$ and $3.$  You can work out the chances of the sum on two independent dice, which ranges from $1$ through $6.$ Tabulate these values $X,$ their chances $p(X),$ and then add a column for $5x^3+8.$ 
$$\begin{array}{rrr}
X & p(X) & 5X^3 + 8 \\
\hline
2 & \frac{1}{9} & 48 \\
3 & \frac{2}{9} & 143 \\
4 & \frac{3}{9} & 328 \\
5 & \frac{2}{9} & 633 \\
6 & \frac{1}{9} & 1088
\end{array}$$
Thus
$$\eqalign{
\mu &= E[5X^3+8] \\&= (5(2)^3+8) \Pr(X=2) + (5(3)^3+8) \Pr(X=3) + \cdots + (5(6)^3+8)\Pr(X=6) \\
&= 408}.$$
Use this to create a new column of the differences between each value of $5X^3+8$ and their expectation, and then a fifth column of their squares:
$$\begin{array}{rrrcc}
X & p(X) & 5X^3 + 8 & 5X^3 + 8 - \mu & (5X^3 + 8-\mu)^2\\
\hline
2 & \frac{1}{9} & 48 & -360& 129600 \\
3 & \frac{2}{9} & 143 & -265 &70225 \\
4 & \frac{3}{9} & 328 & \vdots&\vdots\\
5 & \frac{2}{9} & 633 & \vdots&\vdots\\
6 & \frac{1}{9} & 1088 & 680 & 462400
\end{array}$$
Now find the expectation of the last column using the same formula as before: multiply each value by its probability and add them up.  This is the variance.
It's pretty cumbersome, isn't it?  You can get the correct answer this way for the six-sided dice (with about twice as work involving much larger numbers), but if you know enough algebra you can identify the patterns and simplify the work.  If you have some basic programming skills you can create a small function to do all the calculations for you.  
One conclusion is clear: this is not a good exercise for helping you understand expectations and variances; it's mainly a nuisance question that requires your arithmetical skills and patience.  That's why this kind of question has largely disappeared from statistical pedagogy over the last 40 years.  If you're self-learning from an old textbook, then consider adopting a more recent one.
Finally, writing $E[X]=\bar{X}= 4,$ you will discover that the expectation of $5X^3+8$ (equal to about $408$) is not $5\bar{X}^3+8$ (which is only $328$). In this case, There is no rule of computing with expectations that relates these quantities, but there is an inequality (Jensen's Inequality) that implies the first value cannot be less than the second.
A: $X^3=[(X-\mu)+\mu]^3=(X-\mu)^3+3(X-\mu)^2.\mu+3(X-\mu).\mu^2+\mu^3$
Hence $E(X^3) = E(X-\mu)^3+3\mu E(X-\mu)^2+3\mu^2 E(X-\mu)+\mu^3=3\mu \sigma^2+\mu^3$.
(The 1st term of the four is $0$ because of symmetry + finite support, the 3rd because $E(X)=\mu$)
Now $\mu=7$ and $\sigma^2=\frac{35}{6}$, as you have already found, so you can get $E(X^3)$ directly from them. From there you just use linearity of expectation to get $E(5X^3+8)$.
