Prove that 1 is a triple root of a polynomial I'm studying for an exam and trying to prove whether 1 is a triple root for the polynomial:
$$x^{2n+1}-(2n+1)x^{n+1}+(2n+1)x^n-1$$
for every $n\geq1$.
In our math class we never solved such a problem. So far we only used horner's scheme to prove that someone is a root, double root or triple root.
Can you please help me solve this problem?

$$\begin{align*}
x^{2n+1}&-(2n+1)x^{n+1}+(2n+1)x^n-1\\
&=x^{2n+1}-1-(2n+1)[ x^{n+1}- x^n]\\
&=(x-1)(x^{2n}+x^{2n-1}+\cdots+1)-(2n+1)(x-1)x^n\\
&=(x-1)(x^{2n}+x^{2n-1}+\cdots+x^{n+1}+x^{n-1}+\cdots+1-2nx^n)\\
&=(x-1)[x^n(x^n-1)+x^{n-1}(x^n-x)+\cdots+x^n(x-1)\\
&\qquad\qquad+x^{n-1}(1-x)+x^{n-2}(1-x^2)+\cdots+x(1-x^{n-1})+1-x^n]
\end{align*}$$
Combining corresponding terms from opposite ends of the expression inside the square bracket we have:
$$
=(x-1)[(x^n-1)^2+x(x^{n-1}-1)^2+...+x^{n-1}(x-1)^2]
$$
It is clear that each term inside the square bracket is divisible by $(x-1)^2$ so the whole expression is divisible by $(x-1)^3$
