prove that $(a-b)x^n+(a^2-b^2)x^{n-1}+(a^3-b^3)x^{n-2}+ \dotsm +(a^n-b^n)x+a^{n+1}-b^{n+1}=0$ has at most one real root. 
prove that $$
\left( \text{a}-\text{b} \right) \text{x}^{\text{n}}+\left( \text{a}^2-\text{b}^2 \right) \text{x}^{\text{n}-1}+...+\left( \text{a}^{\text{n}}-\text{b}^{\text{n}} \right) \text{x}+\text{a}^{\text{n}+1}-\text{b}^{\text{n}+1}=0
$$has at most one real root.

I tried to find a nice expression of it, and tried differentiation but failed. How to simplify (It seems like this question can be simplified) it, or directly solve it?
 A: $x$ is a root of this polynomial iff
$$\sum_{r=0}^{n+1}x^ra^{n+1-r}=\sum_{r=0}^{n+1}x^rb^{n+1-r}$$
Multiply $(x-a)(x-b)$ on both sides and get
$$(x-b)(x^{n+2}-a^{n+2})=(x-a)(x^{n+2}-b^{n+2})$$
$$\Longrightarrow (b-a)x^{n+2}-(b^{n+2}-a^{n+2})x+ab(b^{n+1}-a^{n+1})=0$$
Let $P(x)$ be this polynomial, then $P'(x)$ has atmost two roots (if $n$ is odd) and hence $P(x)$ has atmost three distinct roots. But roots $a$ and $b$ are roots of $P(x)$ which were added later when we multiplied by $(x-a)(x-b)$. Hence, original polynomial has atmost one real root. 
A: HINT.- Putting $X=\dfrac xa$ and $Y=\dfrac xb$ we have if $x$ is a real root then
$$a^{n+1}P_n(X)=b^{n+1}P_n(Y)$$ where $P_n(X)$ is the cyclotomic polynomial of degree $n$ when $n+1$ is prime and in general $$P_n(z)=\frac{z^{n+1}-1}{z-1}$$
A: Without loss of generality, assume $a\lt b$. For $x\not\in\{a,b\}$, the equation is equivalent to
$$
\frac{x^{n+2}-a^{n+2}}{x-a}-\frac{x^{n+2}-b^{n+2}}{x-b}=0\tag1
$$
Simple manipulation gives that $(1)$ is equivalent to
$$
\underbrace{\frac{x^{n+2}-a^{n+2}}{b^{n+2}-a^{n+2}}}_{f(x)}=\underbrace{\ \ \frac{x-a}{b-a}\ \ }_{g(x)}\tag2
$$
Note that $f(a)=g(a)=0$ and $f(b)=g(b)=1$. Furthermore, $g(x)$ is linear with slope $\frac1{b-a}$.
If $n$ is even, then $f(x)$ is strictly convex for all $x$. Thus, there is at most one point where $f'(x)=\frac1{b-a}$.
If $n$ is odd, then $f(x)$ is strictly convex for $x\gt0$ and strictly concave for $x\lt0$. Thus, there is at most one $x\ge0$ where $f'(x)=\frac1{b-a}$ and at most one $x\le0$ where $f'(x)=\frac1{b-a}$.
By the Mean Value Theorem, between any two points where $f(x)=g(x)$, there must be a point where $f'(x)=\frac1{b-a}$.
Thus, if $n$ is even, the only points where $f(x)=g(x)$ are $x=a$ and $x=b$. Thus, the original equation has no roots.
If $n$ is odd, there can only be three points where $f(x)=g(x)$, $x=a$ and $x=b$ and possibly one other. Thus, the original equation has at most one root.
