The sum of $3$ real numbers is known to be zero. If the sum of their cubes is $e^{\pi}$ , what is their product equal to?

The answer for this question is $$\frac{e^{\pi}}{3}$$, and I don't understand why.

I tried to let the three real numbers be $$a, b$$, and $$c$$. This meant that \begin{align} a + b + c &= 0 \\ a^3 + b^3 + c^3 &= e^{\pi}\end{align}

How do we get the value of $$abc$$ from the two equations above? I tried cubing the first equation but there are a lot of other terms in the expansion that seem to be unnecessary. Any help would be greatly appreciated.

• math.stackexchange.com/questions/475354/… Apr 6, 2021 at 5:17
• This is famous theorem from 8th grade. Under the conditions of question $a^3+b^3+c^3=3abc$. Apr 6, 2021 at 5:19
• $e^\pi$ is thrown there just for intimidation. Minus points for the question setter. Apr 6, 2021 at 5:20
• @ParamanandSingh I agree that it was chosen partly to intimidate. On the other hand, if this contest is graded based on answers alone (for example, a multiple choice), the designers of this type of questions want to avoid solutions by guessing a collection of values of $a,b,c$ that fit. After all, under those circumstances a contestant can rightfully assume (or guess) that the answer does not depend on the exact values of $a,b,c$. So replacing $e^\pi$ with $18=3^3+(-1)^2+(-2)^3$ would be worse. Apr 6, 2021 at 6:16
• @JyrkiLahtonen: I understand your point! I don't like multiple choice contests for the same reason, but unfortunately they have become more common because the answer sheets can be evaluated using computers with almost no cost. Apr 6, 2021 at 6:36

Using $$c = -(a+b),$$
$$e^\pi = a^3+b^3-(a+b)^3 = -3(a^2 b+b^2 a)=-3ab(a+b) = 3abc$$