Let's represent the solution set of the equation

$$
x^{5}-10 x^{3} y^{2}+5 x y^{4}=0
$$

in the Cartesian coordinate system.
Solution. Factorize the polynomial $x^{5}-10 x^{3} y^{2}+5 x y^{4}$. By factoring out $x$, we get:

$$
x\left(x^{4}-10 x^{2} y^{2}+5 y^{4}\right)=0
$$

A product is zero if and only if one of its factors is zero.

If $x=0$, then the solution set is the line defined by the equation $x=0$, which is the $y$-axis. The other factor of the product is a fourth-degree polynomial. We get a quadratic equation in $y^{2}$, whose solutions are

$$
y^{2}=\frac{10 x^{2} \pm \sqrt{100 x^{4}-20 x^{4}}}{10}=\frac{10 x^{2} \pm \sqrt{80} x^{2}}{10}
$$

From this, $y= \pm \sqrt{1 \pm \frac{2 \sqrt{5}}{5}} x$, so the solution set consists of four lines.

Thus, the solution set of the equation in the Cartesian coordinate system is the following 5 lines:

$$
\begin{array}{rlrl}
x & =0 \\
y_{1} & =\sqrt{1+\frac{2 \sqrt{5}}{5}} x, & y_{2}=\sqrt{1-\frac{2 \sqrt{5}}{5}} x \\
y_{3} & =-\sqrt{1+\frac{2 \sqrt{5}}{5}} x, & y_{4} & =-\sqrt{1-\frac{2 \sqrt{5}}{5}} x
\end{array}
$$

The last four lines pass through the origin and are symmetric with respect to the $y$-axis, dividing the full angle into 10 equal parts.

![](https://cdn.mathpix.com/cropped/2024_05_02_3179dded6a045a73ee39g-1.jpg?height=553&width=415&top_left_y=1244&top_left_x=844)