4. Find all values of the parameter $a$ for which the system

$$
\left\{\begin{array}{c}
\left(a\left|y^{3}\right|+a\left|x^{3}\right|-8\right)\left(x^{6}+y^{6}-3 a^{2}\right)=0 \\
\sqrt{x^{6} y^{6}}=a
\end{array}\right.
$$

has eight distinct solutions. (16 points)
Solution. Let's make a change of variables: $x^{3}=u, y^{3}=v$. Variables $u$ and $v$ can take any values, and $x$ and $y$ are uniquely determined. Then we get

$$
\left\{\begin{array} { c } 
{ ( a | v | + a | u | - 8 ) ( u ^ { 2 } + v ^ { 2 } - 3 a ^ { 2 } ) = 0 } \\
{ | u v | = a , a > 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\left\{\begin{array}{l}
|v|+|u|=8 / a \\
|u v|=a, a>0
\end{array}\right. \\
\left\{\begin{array}{c}
u^{2}+v^{2}=3 a^{2} \\
|u v|=a, a>0
\end{array}\right.
\end{array}\right.\right.
$$

I. $\left\{\begin{array}{l}|v|+|u|=8 / a, \\ |u v|=a, a>0\end{array}\right.$

1) The system has 4 different solutions if

$$
\frac{8}{a}=2 \sqrt{a}, a=2 \sqrt[3]{2}
$$

2) The system has 8 different solutions if

$$
\frac{8}{a}>2 \sqrt{a}, \quad 0 < a < 2 \sqrt[3]{2}
$$

II. $\left\{\begin{array}{l}u^{2}+v^{2}=3 a^{2}, \\ |u v|=a, a>0\end{array}\right.$

1) The system has 4 different solutions if

$$
\sqrt{3} a=\sqrt{2 a}, \quad a=2 / 3
$$

2) The system has 8 different solutions if $\sqrt{3} a>\sqrt{2 a}, a>2 / 3$.

Possible coincidences of solutions in cases I and II:

$$
\left\{\begin{array}{l}
|v|+|u|=8 / a \\
u^{2}+v^{2}=3 a^{2} \\
|u v|=a, a>0
\end{array}\right.
$$

By symmetry, we can consider only the case where $u>0, v>0$. Then

$$
\left\{\begin{array} { c } 
{ v + u = 8 / a , } \\
{ u ^ { 2 } + v ^ { 2 } = 3 a ^ { 2 } , \Leftrightarrow } \\
{ u v = a , a > 0 . }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
v+u=8 / a \\
(u+v)^{2}=3 a^{2}+2 a \\
u v=a, a>0
\end{array}\right.\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-13.jpg?height=1459&width=748&top_left_y=401&top_left_x=1225)

We have $64 / a^{2}=3 a^{2}+2 a, \quad 3 a^{4}+2 a^{3}-64=0$,

$$
a=2,3 a^{4}+2 a^{3}-64=(a-2)\left(3 a^{3}+8 a^{2}+16 a+32\right) . \text { The polynomial } 3 a^{3}+8 a^{2}+16 a+32
$$

has no positive roots. When $a=2$, the solutions in cases I and II coincide, and the system will have 8 solutions.

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-13.jpg?height=491&width=1588&top_left_y=2122&top_left_x=200)

Answer: $a \in(0 ; 2 / 3) \cup\{2\} \cup(2 \sqrt[3]{2} ;+\infty)$.