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.

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 $(a-b)^2,(b-c)^2,(c-a)^2>0$  
