$X$ is the sum of $n$ identical Bernoulli random variables, each with expected value $p$. In other words, $X_{1}, \ldots, X_{n}$ are identical (and independent) Bernoulli random variables with parameter $p$, then $X=X_{1}+\cdots+X_{n}$ and $$ \mathrm{E}[X]=\mathrm{E}\left[X_{1}+\cdots+X_{n}\right]=\mathrm{E}\left[X_{1}\right]+\cdots+\mathrm{E}\left[X_{n}\right]=p+\cdots+p=n p. $$ The variance is: $$ \operatorname{Var}(X)=n p(1-p). $$
The third central moment, defined as $$\mu_{3}=\mathrm{E}\left[(X-\mathrm{E}[X])^{3}\right]=n p(1-p)(1-2 p).$$ I saw this result when I was reading Wikipedia. I know the proof process of expectation and variance, but I don't understand how the result of this third moment is proved?
It is not difficult expanding your expression and using the simple moments to get the final result.
$$ \bbox[5px,border:2px solid black] { \mu_3=\mathbb{E}[X^3]-3\mu\mathbb{E}[X^2]+3\mu^2\mathbb{E}[X]-\mu^3 \qquad (1) } $$
The only "difficult" calculation is $\mathbb{E}[X^3]$ but it can be easy derived using MGF and its properties
EDIT
If you do not want to use MGF, you have to use the definition of simple moments. In order to get the 3rd simple moment you can factorize $x^3$ in the following way
$$x^3=x(x-1)(x-2)+3x^2-2x$$
Thus (I set $q=1-p$ to simplify the notation)
$$\begin{align} \mathbb{E}[X^3] & =\sum_{x=0}^n x(x-1)(x-2)\frac{n!}{x!(n-x)!}p^x q^{n-x}+3npq+3n^2p^2-2np\\ & = p^3n(n-1)(n-2)\sum_{x-3=0}^{n-3}\binom{n-3}{x-3}p^{x-3}q^{n-x}+3npq+3n^2p^2-2np\\ & = p^3n(n-1)(n-2)\underbrace{\sum_{y=0}^{n-3}\binom{n-3}{y}p^{y}q^{(n-3)-y}}_{=1}+3npq+3n^2p^2-2np\\ & = p^3n(n-1)(n-2)+3npq+3n^2p^2-2np \end{align}$$
Now, considering that
$$\mu=\mathbb{E}[X]=np$$
$$\mathbb{E}[X^2]=npq+n^2p^2$$
substitute $\mathbb{E}[X^3]$ in (1) and get the desired result
In fact, after some boring calculations, you get
$$\mu_3=\mathbb{E}[X-\mu]^3=np(1+2p^2-3p)=np(1-p)(1-2p)$$