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Suppose that we have a solution of distinct reals $(a,b,c)$. Let $S = a^2 + b^2 + c^2$.
Then, we have $a^3 + 3a^2 + 25 = 3 (a^2 + b^2 + c^2 ) = 3S$, and similar equations for $b$ and $c$.

Since these are distinct reals, we know that the roots to the cubic equation $X^3 + 3X^2 + 25 - 3S = 0$ must be $a, b, c$.
Hence, by vieta's formula, this tells us $a+b+c = -3, ab+bc+ca = 0$.

Thus, $S = (a^2+b^2+c^2) = (a+b+c)^2 - 2(ab+bc+ca) = 9$. Finally, using the constant coefficient, $abc = - ( 25 - 3S ) = - (25 - 27) = 2$.

Note: We can now solve $x^3 + 3x^2 -2 = 0$ to obtain that $(a,b,c) = ( -1, -1 - \sqrt{3}, -1 + \sqrt{3} )$.

Hint: What can we say about $a^3 + 3a^2 + 25$?

The next hint is in white text, so you will need to Toggle Latex.

Hint: For a given solution set, consider $\color{#FFFFFF} { f(x) = x^3 + 3x^2 + 25 - 3(a^2+b^2+c^2). \text{What are the roots?}}$

first subtract the equation simultaneously to get $\begin{cases} a^3-b^3=-3(a^2-b^2)\\ b^3-c^3=-3(b^2-c^2)\\ c^3-a^3=-3(c^2-a^2) \end{cases}$ since all variable are distinct we can divide both side by (a-b),(b-c),(c-a) respectively. we get $\begin{cases} a^2+ab+b^2+3(a+b)=0\\ b^2+bc+c^2+3(b+c)=0\\ c^2+ca+c^2+3(c+a)=0\\ \end{cases}$ subtract any two equation to get $a+b+c=-3$ add all the equations to get $2(a^2+b^2+c^2)+ab+bc+ca+6(a+b+c)=0$ using the fact$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=9-2(ab+bc+ca)$ $ab+bc+ca=0$ adding all the original equations $a^3+b^3+c^3=6(a^2+b^2+c^2)-75=-21$ using the identity $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=-27+3abc$ so we have $-27+3abc=-21\to abc=2$

The answer is $abc=2$

This is from NMTC 2015 lvl 2

What is the sum of 1!+2!+3!+4!+.....+n!