System 2

The system is equivalent to $\begin{cases} x^3+3y^3-xy^2=1 \\ x^4+6y^4=(x^3+3y^3-xy^2)x+2y\end{cases}$ $\Leftrightarrow\begin{cases} x^3+3y^3-xy^2=1 \\ 6y^4-3y^3x+x^2y^2-2y=0\end{cases}$ $\Leftrightarrow\begin{cases} x^3+3y^3-xy^2=1 \\ y(6y^3-3y^2x+x^2y-2)=0\end{cases}$ From the second equation

$*y=0$ we have $x^3=1$ $\therefore (x;y)=(1;0)$

$*6y^3-3y^2x+x^2y-2=0$ , set $x=ky \ (k\in R)$ , the system becomes $\Leftrightarrow\begin{cases} k^3y^3+3y^3-ky^3=1 \\ 6y^3-3ky^3+k^2y^3=2\end{cases}$ $\Leftrightarrow\begin{cases} y^3=\frac{1}{k^3-k+3}\\y^3=\frac{2}{k^2-3k+6}\end{cases}$ $\Leftrightarrow\begin{cases}\frac{1}{k^3-k+3}=\frac{2}{k^2-3k+6} \\ y^3=\frac{1}{k^3-k+3}\end{cases}$ $\Leftrightarrow\begin{cases}2k^3-k^2+k=0\\ y^3=\frac{1}{k^3-k+3}\end{cases}$ $\Leftrightarrow\begin{cases} k=0 \\ y=\frac{\sqrt[3]{9}}{3}\end{cases}$ $\Rightarrow (x;y)=\bigg(0;\frac{\sqrt[3]{9}}{3}\bigg)$ So $\displaystyle\sum_{m=1}^{n} x_m+y_m = 1+\frac{\sqrt[3]{9}}{3}\approx 1.69$