Deduce that $\mathbb E(X^3)=1^3+2^3+3^3+4^3+5^3+6^3$ A fair die is tossed  and let  the random  variable $X$ be the number that appears.
Deduce that 
$$
\mathbb E(X^3)=\frac{1^3+2^3+3^3+4^3+5^3+6^3}6.
$$
First of all, I would like to know the probability distribution  of this random variable $X$.
 A: $X$ takes the values $1$, $2$, $3$ ,$4$, $5$, and $6$ (assuming a six-sided die). Since the die is fair, outcomes are equally likely; so $X$ takes the value $i$ with probability $1/6$ for each $i=1, 2,3,4,5,6$. 
The probability distribution of $X$ is therefore
$$
p_X(i)=\textstyle{1\over 6},\quad i=1, 2, \ldots, 6.
$$
As a warm up to your problem, let's find $\Bbb E(X)$. The expected value of $X$  is
$$
\Bbb E(X)=\sum_{i=1}^6 \,i\, p_X(i)= 1\cdot {1\over6}+2\cdot {1\over6}+3\cdot {1\over6}+4\cdot {1\over6}
+5\cdot {1\over6}+6 \cdot {1\over6}=3.5.
$$
To find the expected value of a function of $X$, you could use the following fact:
Fact:
For a discrete variable $X$ that takes the values $x_1, x_2,\ldots,x_n$, 
the expectation of a function $h(X)$ of $X$ is
$$\Bbb E\bigl(h(X)\bigr) =\sum_{i=1}^n h(x_i) p_X(x_i).$$
To find $\Bbb E(X^3)$, we apply the above fact with   $h(x)=x^3$:
$$
\Bbb E( X^3)=\sum_{i=1}^6\, i^3 p_X(i)= ( 1)^3\cdot {1\over6}+( 2)^3\cdot {1\over6}+( 3)^3\cdot {1\over6}+( 4)^3\cdot {1\over6}
+( 5)^3\cdot {1\over6}+( 6)^3 \cdot {1\over6} .
$$
