Solve the following system of equations:

$$
x^{3}+y^{3}=a, \quad x y(x+y)=b
$$
Add to the first equation three times the second equation, so that the left side becomes $(x+y)^{3}$. From this, by taking the cube root,

$$
x+y=\sqrt[3]{a+3 b}
$$

then substituting this into the second equation and dividing,

$$
x y=\frac{b}{\sqrt[3]{a+3 b}}
$$

provided that $a+3 b \neq 0$. Let's confine ourselves to this case for now. Under our condition, $x+y$ and $x y$ always exist and are uniquely determined values, because the cube root is always performable and unique in the realm of real numbers. Let

$$
\sqrt[3]{a+3 b}=B \quad \text { and } \quad b / B=C
$$

then $x$ and $y$ are the two roots of the following equation:

$$
z^{2}-B z+C=0
$$

that is,

$$
\frac{1}{2}\left(B+\sqrt{B^{2}-4 C}\right) \quad \text { and } \quad \frac{1}{2}\left(B-\sqrt{B^{2}-4 C}\right)
$$

in any order. The discriminant, expressed in terms of $a$ and $b$,

$$
B^{2}-4 C=\frac{1}{B}\left(B^{3}-4 B C\right)=\frac{1}{\sqrt[3]{a+3 b}}(a+3 b-4 b)=\frac{a-b}{\sqrt[3]{a+3 b}}
$$

thus the roots are

$$
\frac{1}{2}\left(\sqrt[3]{a+3 b}+\sqrt{\frac{a-b}{\sqrt[3]{a+3 b}}}\right) \quad \text { and } \frac{1}{2}\left(\sqrt[3]{a+3 b}-\sqrt{\frac{a-b}{\sqrt[3]{a+3 b}}}\right)
$$

$x$ and $y$ are real if the number under the square root is positive or zero. The value of the fraction under the square root is positive if the numerator and denominator are both positive or both negative, that is, if

$$
a-b>0 \quad \text { and } \quad a+3 b>0 \quad \text { or } \quad a-b<0 \quad \text { and } \quad a+3 b<0
$$

If $b>0$, then $a+3 b=a-b+4 b>a-b$, so it is enough to require that $a-b>0$; if, however, $b \leq 0$, then $a-b=a+3 b-4 b \geq a+3 b$, and thus it is enough to require that the latter expression is positive. Applying similar considerations in the second case, we get that two different real values are given for $x$ and $y$ (and their roles can be interchanged), if

$$
\begin{aligned}
& \text { I. } a-b>0 \quad \text { and } \quad b>0, \quad \text { or } \\
& \text { II. } a+3 b>0 \quad \text { and } \quad b \leq 0, \quad \text { or } \\
& \text { III. } a+3 b<0 \quad \text { and } \quad b \geq 0, \quad \text { or } \\
& \text { IV. } a-b<0 \quad \text { and } b<0 \text {. }
\end{aligned}
$$

The value of the square root is zero if $a=b$ (and $\neq 0$, otherwise $a+3 b=0$ would hold), in which case there is one solution:

$$
x=y=\frac{1}{2} \sqrt[3]{a+3 b}
$$

If finally $a+3 b=0$, then $x+y=0$, and this is only consistent with the second equation if $b=0$; in this case $a=-3 b=0$ also holds. Then for any $x$, choosing $y=-x$ gives a solution to the system of equations. - If, however, $a+3 b=0$ and $b \neq 0$, then the system of equations has no solution.

Remarks. 1. If in (2) we take the square root of the numerator and denominator separately, we may lose the III. and IV. cases.

2. Using the identity $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$ and dividing the two equations $(x, y, x+y \neq 0) \mathrm{the}$

$$
\frac{x^{2}-x y+y^{2}}{x y}=\frac{a}{b}
$$

equation, and then by the substitution $x / y=u$ and rearranging, we can arrive at the quadratic equation

$$
u^{2}-(1+a / b) u+1=0
$$

Calculating $u$, the one root can be expressed in terms of the other. Substituting this expression, for example, into (1), the roots themselves can also be calculated, but they appear in a significantly more complicated form than the above.