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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.

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  • $\begingroup$ math.stackexchange.com/questions/475354/… $\endgroup$ Apr 6, 2021 at 5:17
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    $\begingroup$ This is famous theorem from 8th grade. Under the conditions of question $a^3+b^3+c^3=3abc$. $\endgroup$
    – Paramanand Singh
    Apr 6, 2021 at 5:19
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    $\begingroup$ $e^\pi$ is thrown there just for intimidation. Minus points for the question setter. $\endgroup$
    – Paramanand Singh
    Apr 6, 2021 at 5:20
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    $\begingroup$ @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. $\endgroup$ Apr 6, 2021 at 6:16
  • $\begingroup$ @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. $\endgroup$
    – Paramanand Singh
    Apr 6, 2021 at 6:36

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Using $c = -(a+b),$

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

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