Find the summation of given expression: $\sum_{k=1}^n k^3\binom nk$

Question

I am trying to solve the following question which is Ex3 from Arthur Engel, Problem Solving strategies. Here is the question: $\sum_{k=1}^n k^3 {n \choose k}$ and asks to find the sum. I am sincerely having a hard time trying to figure out how to solve it and get $n^2 (n+3)(2)^{n-3}$. I would really appreciate if someone could show the step-by-step process. Thank you!

Answer 1

Expanding on @LordSharktheUnknown's hint (since a duplicate of this question was recently posted), apply $x\frac{d}{dx}$ thrice to $\sum_{k=1}^n\binom{n}{k}x^k=(1+x)^n$. First,$$\sum_k\binom{n}{k}kx^k=nx(1+x)^{n-1}=n(1+x)^n-n(1+x)^{n-1}.$$Second,$$\begin{align}\sum_k\binom{n}{k}k^2x^k&=n^2x(1+x)^{n-1}-n(n-1)x(1+x)^{n-2}\\&=n^2(1+x)^n-(2n^2-n)(1+x)^{n-1}+n(n-1)(1+x)^{n-2}.\end{align}$$Third,$$\sum_k\binom{n}{k}k^3x^k=n^3x(1+x)^{n-1}-(2n^2-n)(n-1)x(1+x)^{n-2}+n(n-1)(n-2)x(1+x)^{n-3}.$$The case $x=1$ simplifies to $\sum_k\binom{n}{k}k^3=n^2(n+3)2^{n-3}$.