10.4. Real numbers ${ }^{a}$ and ${ }^{b}$ are such that $a^{3}+b^{3}=1-3 a b$. Find all values that the sum ${ }^{a+b}$ can take.
Answer: -2 and 1.

Solution. Let $a+b=x$ and $a b=y$, then the equation from the condition can be written as $x^{3}-3 x y=1-3 y$, from which $x^{3}-1=3 x y-3 y=3 y(x-1)$, we easily obtain the first possible value $a+b=x=1$. Dividing by the factor $x-1$, we have $x^{2}+x+1-3 y=0$. Substituting $x=a+b$ and $y=a b$ here, we get $a^{2}-(b-1) a+b^{2}+b+1=0$. Considering the last equation as a quadratic in ${ }^{a}$, we find its discriminant, equal to $-3(b+1)^{2} \leq 0$. This equation is solvable in the unique case $b=-1$ and its only root will be $a=-1$. In this case, $a+b=-2$ - the second possible answer to the problem. Verification of the found values is not required, as the transformations we performed were equivalent. Grading criteria. Finding ${ }^{a+b=1}: 3$ points. Finding ${ }^{a+b=-2}: 4$ points. If both answers are guessed with examples of ${ }^{a}$ and $^{b}$ on which they are achieved: 1 point.