Calculate $\mathbb{E}[X^3]=\mathbb{M}_3[X]$ with $X \sim B(n,p).$
This formula we get in the lecture:
$$\mathbb{E}[X^3]=\sum\limits_{x=0}^{n}x^3 \binom{n}{x}p^x(1-p)^{n-x}=q\frac{\mathrm{d}}{\mathrm{d}q}\sum\limits_{x=0}^{n} \binom{n}{x}q^x(1-p)^{n-x}\bigg|_{q=p}=\left(q\frac{\mathrm{d}}{\mathrm{dq}}\right)^3 (q+(1-p)^n\bigg|_{q=p}$$
I derivate the last term since one hour again and again, but unfortunately I did not get the right expectation.
Is this the right idea? Alternative I used the Laplace-Transformation given by: $$\begin{align}\mathscr{L}_X(\alpha)&=\sum\limits_{x=0}^{\infty}\mathrm{e}^{-\alpha x} p_x^X,\qquad p_x^x = \binom{n}{x}p^x(1-p)^{n-x}, \alpha \ge 0 \\ \mathbb{M}_p[X] &= (-1)^p \frac{\mathrm{d}}{\mathrm{d}\alpha} \mathscr{L}_X(\alpha)\bigg|_{\alpha=0}\end{align}$$
Any hints for me?
First I know you requested no MGF but the useful observation here is that considering $X \sim \mbox{Bin}(n,p)$ is the same as considering $$ \begin{align} X = \sum_{i=1}^{n}B_i, \end{align} $$ where $B_i$ are independent Bernoulli random variables with $P(B_i = 1) = p$, and so knowing the moment generating function of a Bernoulli random variable we can easily calculate the moment generating function of a Binomial random variable.
Anyway without the MGF and working with the binomial coefficients themselves you can drop the power like this $$ \begin{align} \mathbb{E}\left[ X^3 \right] &= \sum_{x=0}^{n} x^3 \begin{pmatrix} n \\k\end{pmatrix}p^x(1-p)^{n-x} \\ &= np\sum_{x=1}^{n}x^2\begin{pmatrix} n-1 \\x-1\end{pmatrix}p^{x-1}(1-p)^{(n-1)-(x-1)} \\ &= np\sum_{y=0}^{m}(y+1)^2 \begin{pmatrix} m \\y\end{pmatrix}p^y(1-p)^{m-y} &\mbox{where }y=x-1, m=n-1\\ &= np\mathbb{E}\left[ (Y+1)^2\right], &\mbox{where }Y \sim \mbox{Bin}(n-1,p). \end{align} $$ That new expectation is then easier to work with. As an additional comment I am confused about that expression you have taking the derivative with respect to $q$ and so on.