Let α be a solution of x^2+x+1=0, and for some a and b in R, [4, a, b] [1, 16, 13; -1, -1, 2; -2, -14, -8] = [0, 0, 0]. If 4/(α^4)+m/(α^a)+n/(α^b)=3,…

Let $\displaystyle \alpha$ be a solution of $\displaystyle x^2+x+1=0$, and for some $\displaystyle a$ and $\displaystyle b$ in $\displaystyle \mathbb{R},\left[\begin{array}{lll}4 & a & b\end{array}\right]\left[\begin{array}{ccc}1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$. If $\displaystyle \frac{4}{\alpha^4}+\frac{m}{\alpha^a}+\frac{n}{\alpha^b}=3$, then $\displaystyle m+n$ is equal to $\displaystyle \_\_\_\_$