Morning coffe

$\begin{aligned} (a+b)^2 & = a^2 + 2ab + b^2 \\ 1 & = 2 + 2ab \\ \implies ab & = - \frac 12 \end{aligned}$

$\begin{aligned} (a-b)^2 & = a^2 - 2ab + b^2 \\ & = 2 + 1 \\ \implies a-b & = \sqrt 3 & \small \color{#3D99F6} \text{for }a>b \end{aligned}$

Now, we have:

$\begin{aligned} \frac {a^3+b^3}{a^3-b^3} & = \frac {(a+b)(a^2-ab+b^2)}{(a-b)(a^2+ab+b^2)} \\ & = \frac {(1)\left(2+\frac 12\right)}{\sqrt 3\left(2-\frac 12\right)} \\ & = \frac {\frac 52}{\frac {3\sqrt 3}2} = \frac 5{3\sqrt 3} = \frac {5\sqrt 3}9 \end{aligned}$

$\implies m+n = 5+9 = \boxed{14}$ .