67. (8) A polynomial $f(x)$ of degree $2n-1$ with real coefficients is such that the polynomial $(f(x))^{2}-f(x)$ is divisible by the polynomial $\left(x^{2}-x\right)^{n}$. What can the leading coefficient of the polynomial $f(x)$ be?
$\square$ Answer: $\pm \frac{(2 n-2)!}{(n-1)!}$.

Since the product $f(x)(f(x)-1)$ is divisible by $x^{n}(x-1)^{n}$, and one of the factors $f(x)$ and $f(x)-1$ is not divisible by $x$, then one of them is divisible by $x^{n}$. Similarly, one of them is divisible by $(x-1)^{n}$, and the other is not, otherwise the degree of $f(x)$ would be at least $2 n$. Without loss of generality, $f(x)$ is divisible by $(x-1)^{n}$, and $f(x)-1$ is divisible by $x^{n}$ (otherwise, replace $f(x)$ with the polynomial $1-f(x)$). Note that if there are two such polynomials $f(x)$ and $g(x)$ of degree less than $2 n$, then their difference $f(x)-g(x)$ is divisible by both $x^{n}$ and $(x-1)^{n}$, which is possible only if $f(x)-g(x) \equiv 0$. Thus, there is at most one polynomial $f(x)$ with the required properties, so it remains to present one and compute its leading coefficient (the leading coefficient of the polynomial $1-f(x)$ will have the opposite sign). Let $F(t, k)=(t+1)(t+2) \ldots(t+k-1) /(k-1)!$. Consider the polynomials $g_{0}(x)=1, g_{k}(x)=F(0, k)+F(1, k) x+\ldots+F(n-1, k) x^{n-1}$ for $k=1, \ldots$ It is not difficult to establish that $(1-x) g_{k}(x) \equiv g_{k-1}(x)\left(\bmod x^{n}\right)$: the equality of coefficients at $x^{t}$ is equivalent to the directly verifiable identity $F(t, k)-F(t-1, k)=F(t, k-1)$. From this, $(1-x)^{n} g_{n}(x) \equiv g_{0}(x)=1\left(\bmod x^{n}\right)$, so the polynomial $f(x)=(1-x)^{n} g_{n}(x)$ satisfies both conditions, it remains to compute its coefficient at $x^{n-1}$: it is equal to $(-1)^{n} F(n-1, n)=$ $(-1)^{n} \frac{(2 n-2)!}{(n-1)!}$.