Unexpected Equality of Finite Sums (proving without actually evaluting both sums)

Question

Consider the following identities

\begin{align*} \sum_{k=0}^n\binom nk x^k(1-x)^{n-k}&=1\tag1\\ \sum_{k=0}^n\frac kn\binom nkx^k(1-x)^{n-k}&=x\tag2\\ \sum_{k=0}^n\frac kn\left(1-\frac nk\right)x^k(1-x)^{n-k}&=x(1-x)\tag3 \end{align*}

There are quite straightforward to prove by first using the Binomial Theorem for $(1)$ and then deducing $(2)$ and $(3)$. However, we can obtain the following chain of equalities

$$\small\sum_{k=0}^n\left(x-\frac kn\right)^2\binom nkx^k(1-x)^{n-k}=\frac1n[x(1-x)]=\sum_{k=0}^n\left(x(1-x)-\frac kn\left(1-\frac kn\right)\right)\binom nkx^k(1-x)^{n-k}$$

In particular, therefore we can deduce that

$$\small\sum_{k=0}^n\left(x-\frac kn\right)^2\binom nkx^k(1-x)^{n-k}=\sum_{k=0}^n\left(x(1-x)-\frac kn\left(1-\frac kn\right)\right)\binom nkx^k(1-x)^{n-k}\tag{$\star$}$$

This is where the fun begins! Is it somehow possible to obtain $(\star)$, without relying on our intermediate chain of equalities? Playing around with the two sums I cannot see a direct way to attack the problem and honestly I do not know what else to do.

Is it possible to show $(\star)$ without actually evaluating both sums?

Thanks in advance!

Answer 1

Looks like I have overlooked something quite trivial as cross-posting the problem on AoPS attracted a simple solution by computation, which I will post here for reference. All credit due to the user ysharifi.

Well, just subtract the two sums (and by the way we don't need the condition $x \in [0,1]$) $$\small\begin{align*} &~~~~~\sum_{k=0}^n\left(\left(x-\frac kn\right)^2-x(1-x)+\frac{k}{n}\left(1-\frac{k}{n}\right)\right)\binom{n}{k}x^k(1-x)^{n-k} \\&=\sum_{k=0}^n\left(2x^2-x-\frac{2k}{n}x+\frac{k}{n}\right)\binom{n}{k}x^k(1-x)^{n-k} \\ &=(2x^2-x)\sum_{k=0}^n\binom{n}{k}x^k(1-x)^{n-k}+(1-2x)\sum_{k=1}^n\frac{k}{n}\binom{n}{k}x^k(1-x)^{n-k}\\ &=2x^2-x+(1-2x)\sum_{k=1}^n\binom{n-1}{k-1}x^k(1-x)^{n-k}=2x^2-x+(1-2x)\sum_{k=0}^{n-1}\binom{n-1}{k}x^{k+1}(1-x)^{n-1-k}\\ &=2x^2-x+(1-2x)x\sum_{k=0}^{n-1}\binom{n-1}{k}x^k(1-x)^{n-1-k}\\ &=2x^2-x+(1-2x)x=0. \end{align*}$$