algebra precalculus - Polynomial with real roots

Consider the polynomial: $$f=X^4+4X^3+6X^2+aX+b$$
We know that $f$ has four real roots. Let $x_1,x_2,x_3,x_4$ be the roots of this polynomial. How can one compute $$x_1^{2015}+x_2^{2015}+x_3^{2015}+x_4^{2015}?$$

If $a=4$ and $b=1$, we obtain a self-reciprocal (palindromic) polynomial. We can write $f=(X+1)^4$, thus $x_1=x_2=x_3=x_4=1$. Hence the sum computes to $-4$.
Are there any other cases to consider ($a,b$)? I thought using the formula for the quartic equation and paying attention to the cases where we have only real roots.
Any ideas? Thank you!

Answer

We know the following to be true:

$$
\tag1x_1+x_2+x_3+x_4 = -4
$$
$$

\tag2x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4 = 6
$$
$$
\tag3x_1 x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2 = -a
$$
$$
\tag4x_1x_2x_3x_4 = b
$$

Now we can square $(1)$ to get

$$
\begin{align}
(x_1+x_2+x_3+x_4)^2 = &x^2_1+x^2_2+x^2_3+x^2_4\\
&+2[x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1+x_3+x_2x_4]
\end{align}
$$

Substituting in values from $(1)$ gives

$$
\begin{align}
(-4)^2 &= x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}+2(6)\\
16&=x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}+12\\
4&=x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}
\end{align}
$$

Using the Cauchy-Schwartz Inequality, we get

$$
\left(x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}\right)(1^2+1^2+1^2+1^2)\geq (x_{1}+x_{2}+x_{3}+x_{4})^2
$$

where equality holds when

$$
\tag5x_{1} = x_{2} = x_{3} = x_{4}
$$

Thus, the equality condition holds, because $4\cdot 4= (-4)^2$. So from $(1)$ and $(5)$, we have $$x_{1} = x_{2} = x_{3} = x_{4}=-1$$

Setting the values of $x_{1} = x_{2} = x_{3} = x_{4}=-1$ in $(3)$ and $(4)$, we get $(a,b)=(4,1)$, which is thus the unique solution. So we can evaluate the sum as

$$
\begin{align}
x^{2015}_{1}+x^{2015}_{2}+x^{2015}_{3}+x^{2015}_{4} &= (-1)+(-1)+(-1)+(-1)\\
&=-4
\end{align}
$$