If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$. If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.
I think it is trivial because they are distinct.
So I wonder just saying "Since they are distinct" is enough to prove it?
Of course there could be several more detailed versions but I just want to know that reasoning is true or not.
 A: HINT:
$$a^3+b^3+c^3-3abc$$
$$=(a+b)^3-3ab(a+b)+c^3-3abc=(a+b)^3+c^3-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2-3ab\}$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$=(a+b+c)\frac{\{(a-b)^2+(b-c)^2+(c-a)^2\}}2$$
which will be $>=<0 $ according as $a+b+c>=<0$ as for real distinct $a,b,c,$ each of $(a-b)^2,(b-c)^2,(c-a)^2>0$  
A: You statement is wrong, see $(0,-1,1)$.
A: $$
(-1)^3 + (-2)^3 + 3^3 = -1 - 8 + 27 = 18 = 3(-1)(-2)(3)
$$
A: Note: I failed to notice that the problem as stated allows $a,b$, and $c$ to be arbitrary real numbers. As others have noted, in this generality the result is false. It is, however, true if $a,b,c\ge 0$. If $-a^3-b^3-c^3+3abc=0$, then 
$$\frac13\left(a^3+b^3+c^3\right)=abc=\sqrt[3]{a^3b^3c^3}\;,$$
i.e., the arithmetic mean of $a^3,b^3$, and $c^3$ is the same as their geometric mean. Now use the full form of the AM-GM inequality.
I’ll leave this up for now in case it turns out that the problem was actually supposed to include the hypothesis that $a,b,c\ge 0$.
