<s>[INST] Are all the roots of $x^n-kx^{n-1}-kx^{n-2}-\cdots-kx-k=0$ "distinct"? Are all the roots of $x^n-kx^{n-1}-kx^{n-2}-\cdots-kx-k=0$ "distinct", where $n,k$ are integers and $n,k\geq 2$? And why?
 [/INST] Write :$$f(x)=x^n-k{x^n-1\over x-1}$$
Say it has double root $a$, then $f(x) = g(x)(x-a)^2$ so we have $$g(x)(x-a)^2(x-1)= \underbrace{x^{n+1}-x^n- kx^n+k}_{p(x)}$$ So $a$ is a root of $p(x)$ and $p'(x)$. Since $$p'(x)= (n+1)x^n-n(k+1)x^{n-1}$$ we have $a=0$ or $a={n(k+1)\over n+1}$ only possible double roots. 
Is this helpful? 
Yes, it is helpful. If $a=\frac{n(k+1)}{n+1}$, then we obtain from $f(a)=0$ that $a^n=\frac{k(n+1)}{k+1}$. However, this contradicts to $a^n
=\left(\frac{n(k+1)}{n+1}\right)^n\geq \left(\frac{2(2+1)}{2+1}\right)^n
=2^n\geq n+1>\frac{k(n+1)}{k+1}$.
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