Let's determine the real roots of the system of equations:

$$
\begin{aligned}
(x+y)^{4} & =6 x^{2} y^{2}-215 \\
x y\left(x^{2}+y^{2}\right) & =-78
\end{aligned}
$$
# I. Solution:

$$
\begin{aligned}
(x+y)^{4} & =6 x^{2} y^{2}-215 \\
x y\left(x^{2}+y^{2}\right) & =-78
\end{aligned}
$$

(2) can also be written as:

$$
x y(x+y)^{2}=2 x^{2} y^{2}-78
$$

From (2), $x y \neq 0$, so we can multiply (1) by $x^{2} y^{2}$ and square (3):

$$
\begin{aligned}
& x^{2} y^{2}(x+y)^{4}=6 x^{4} y^{4}-215 x^{2} y^{2} \\
& x^{2} y^{2}(x+y)^{4}=\left(2 x^{2} y^{2}-78\right)^{2}
\end{aligned}
$$

From (4) and (5), it follows that the two right-hand sides are equal. Let $x^{2} y^{2}=z$ and note that if $x$ and $y$ are real, then $z>0$.

$$
6 z^{2}-215 z=(2 z-78)^{2}=4 z^{2}-312 z+6084
$$

Thus,

$$
2 z^{2}+97 z-6084=0
$$

From which the positive root is

$$
z=36 \text {. }
$$

Therefore,

$$
z=x^{2} y^{2}=36
$$

And since $x y<0$ from (2), we have

$$
x y=-6 \text {. }
$$

From (3),

$$
(x+y)^{2}=\frac{72-78}{-6}=1
$$

Thus,

$$
x+y= \pm 1 \text {. }
$$

From (9), $y=-x \pm 1$; substituting this value into (7),

$$
-x^{2}+x+6=0
$$

or

$$
-x^{2}-x+6=0
$$

From (10), $x_{1}=3, x_{2}=-2$, and from (11), $x_{3}=-3, x_{4}=2$, and thus from (7),

$$
y_{1}=-2, \quad y_{2}=3, \quad y_{3}=2, \quad y_{4}=-3 .
$$

Remark: We can also arrive at the equation under (6) as follows.

Write (1) as follows:

$$
x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4}=6 x^{2} y^{2}-215
$$

Then,

$$
x^{4}+y^{4}+4 x y\left(x^{2}+y^{2}\right)=-215
$$

And thus, considering (2),

$$
x^{4}+y^{4}=-4(-78)-215=312-215=97
$$

Squaring (2),

$$
x^{2} y^{2}\left(x^{4}+y^{4}+2 x^{2} y^{2}\right)=6084
$$

Considering (12) and substituting $x^{2} y^{2}$ with $z$, we arrive at the equation (6).

II. Solution: Let $(x+y)^{2}=u$ and $x y=v$, then $x^{2}+y^{2}=(x+y)^{2}-2 x y=u-2 v$, and our system of equations becomes

$$
\begin{aligned}
u^{2}-6 v^{2}+215 & =0 \\
v(u-2 v)+78 & =u v-2 v^{2}+78=0
\end{aligned}
$$

Note that if $x$ and $y$ are real, then $u \geq 0$. Subtracting (1) from three times (2),

$$
3 u v-u^{2}+19=0,
$$

from which

$$
v=\frac{u^{2}-19}{3 u}
$$

Substituting this value into (1),

$$
u^{2}-6 \frac{u^{4}-38 u^{2}+361}{9 u^{2}}+215=0
$$

Thus,

$$
u^{4}+721 u^{2}-722=0,
$$

and from this, the only positive root is

$$
u=1 . \quad .
$$

From (3),

$$
v=\frac{1-19}{3}=-6
$$

Therefore,

$$
\begin{aligned}
x+y & =\sqrt{u}= \pm 1, \\
x y & =v=-6 .
\end{aligned}
$$

The calculations can proceed as in the first solution.

III. Solution: Using the equation from (12) in the first solution,

$$
x^{4}+y^{4}=97,
$$

and from the previous solutions,

$$
x^{2} y^{2}=36 \text {, } \quad
$$

that is,

$$
x^{4} y^{4}=1296
$$

we see that $x^{4}$ and $y^{4}$ are the two roots of the quadratic equation

$$
t^{2}-97 t+1296=0
$$

These roots are

$$
t_{1}=81, \quad t_{2}=16 \text {. }
$$

Thus,

$$
x^{4}=81 \quad \text { and } y^{4}=16 \text {, or vice versa } x^{4}=16 \text { and } y^{4}=81 \text {, }
$$

from which, considering $x y<0$, the four real roots pairs mentioned in the previous solutions are obtained.