Solve the following system of equations:

$$
\begin{array}{r}
x^{2}-x y+y^{2}=2, \\
x^{3}-y^{3}=4 .
\end{array}
$$
The introduction of new unknowns can reduce our task to solving a system of simpler equations. For example,

$$
x-y=u, \quad \text { and } \quad x y=v
$$

then our equations transform as follows:

$$
\begin{gathered}
x^{2}-x y+y^{2}=(x-y)^{2}+x y=u^{2}+v=2 \\
x^{3}-y^{3}=(x-y)^{3}+3 x y(x-y)=u^{3}+3 u v=4
\end{gathered}
$$

Substituting the expression for $v$ from the first equation into the second:

$$
u^{3}+3 u\left(2-u^{2}\right)=4
$$

After rearranging and simplifying, the left side can be factored:

$$
\begin{gathered}
u^{3}-3 u+2=0 \\
\left(u^{3}-u\right)-2(u-1)=(u-1)[u(u+1)-2]=(u-1)\left(u^{2}+u-2\right)
\end{gathered}
$$

This product is zero if and only if one of its factors is zero, so either $u-1=0, u_{1}=1$, or $u^{2}+u-2=0$. The latter equation has two roots $u_{2}=1$ and $u_{3}=-2$. The corresponding $v$ values can be calculated from equation (1a):

$$
v_{1}=1, \quad v_{2}=1, \quad v_{3}=-2
$$

The $x$ and $y$ values corresponding to the root system $u_{1}=1, v_{1}=1$ can be calculated from the equations

$$
u=x-y=1, \quad v=x y=1
$$

$$
\begin{array}{cc}
x_{1}=\frac{\sqrt{5}+1}{2}, & y_{1}=\frac{\sqrt{5}-1}{2} \\
x_{2}=-\frac{\sqrt{5}-1}{2}, & y_{2}=-\frac{\sqrt{5}+1}{2}
\end{array}
$$

The root pair $u_{2}, v_{2}$ does not yield new solutions. The equation system $x-y=-2, x y=-2$ corresponding to the root pair $u_{3}=-2, v_{3}=-2$ has no real solutions.

![](https://cdn.mathpix.com/cropped/2024_05_02_321c54c8dc7e12013fb7g-1.jpg?height=748&width=814&top_left_y=1605&top_left_x=642)

Remark. Without proof, we mention that the image of (1) in the Cartesian coordinate system is an ellipse whose center is the origin, and whose symmetry axes form a $45^{\circ}$ angle with the coordinate axes, with its major axis passing through the I. and III. quadrants. The image of (2) is a so-called third-degree curve. The pairs of numbers $x, y$ that satisfy both equations are the points of intersection of the curves, in this case, points of tangency, where the curves also share a common tangent.