Represent $\dfrac{\lambda_1^M-\lambda_2^M}{\lambda_1-\lambda_2}$ in terms of $\lambda_1+\lambda_2$ and $\lambda_1\lambda_2$ I have a problem as follows: Let $\lambda_1, \lambda_2$ are roots of the equation $\lambda^2-a\lambda+b=0.$ It can be proved easily (by induction for example) that the quantity $$\dfrac{\lambda_1^M-\lambda_2^M}{\lambda_1-\lambda_2}$$ is a polynomial $P_M$ in $a,b$.
My question is: Can we write down a precise formula for $P_M$?
I found a recursive formula for $P_M$'s which reads $$P_{M+1}=aP_M-bP_{M-1},$$ however it seems to be not enough to answer my question.
Any hint or idea is appreciated.
Thanks a lot!

If $\lambda_1 = re^{i\theta}$, $\lambda_2 = re^{-i\theta}$, then
$$\begin{cases}
a = \lambda_1 + \lambda_2 = 2r\cos\theta\\
b = \lambda_2 \lambda_2 = r^2
\end{cases}
\quad\implies\quad
\begin{cases}
r &= \sqrt{b}\\
\cos\theta &= \frac{a}{2\sqrt{b}}
\end{cases}
$$
For integer $M > 1$,
$$P_M = \frac{\lambda_1^M - \lambda_2^M}{\lambda_1 - \lambda_2} = \frac{r^M\sin(M\theta)}{r\sin\theta} = r^{M-1}U_{M-1}(\cos\theta) = \sqrt{b}^{M-1}U_{M-1}\left(\frac{a}{2\sqrt{b}}\right)$$
where $U_n(x)$ is the Chebyshev polynomials of the second kind. 
This implies $P_M$ is a homogenous polynomial of degree $M-1$ in $a$ and $\sqrt{b}$.
