Rational root coefficient I saw this question in my exam recently,

If a, b, c are distinct rational roots of $x^3+ax^2+bx+c=0$, find the values of a, b, c.

Can someone give me a hint or answer? I tried factoring it and then expand it all out, but it doesn't seem very useful as they are all "rational". Can someone help me with it?

Apologies for the length of the argument.  I tried to format it to be more comprehensible.

Multiplying out $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$, so we have the equations: 
$1)$ $a=-a-b-c$
$2)$ $b=ab+ac+bc$
$3)$ $c=-abc$
Notice that if $a = 0$ or $b = 0$, then by the $(3)$, $c=0$ contradicting the fact that they must be distinct.  Therefore $a,b\neq 0$.  
Now if $c=0$, then by $(1)$ and $(2)$, $2a=-b$ and $b=ab$, which has the solution $a=1, b=-2, c=0$.  The goal now is to show no other solution exists.  
Consider when $c \neq 0$.  Cancelling the $c$'s in $(3)$ gives us $1=-ab \implies a=-\frac{1}{b}$.  Then we can simplify $(1)$ into $b^2+cb-2=0$ 
$\implies b=\frac{-c \pm \sqrt{c^2+8}}{2}$
and $(2)$ simplifies to $(c-1)b^2-b-c=0$ 
$\implies b=\frac{1\pm\sqrt{4c^2-4c+1}}{2(c-1)}=\frac{1\pm(2c-1)}{2(c-1)}$
for $c\neq 1$ 
(if $c=1$, then $(c-1)b^2-b-c=0\implies b=-1 \implies a=1$, contradicting distinct values again).  
Setting the equations for $b$ equal yields 
$\frac{-c \pm \sqrt{c^2+8}}{2}=\frac{1\pm(2c-1)}{2(c-1)}$
or
$\frac{-c \pm \sqrt{c^2+8}}{2}=\frac{1\mp(2c-1)}{2(c-1)}$
Changing these two equations to the one
$c^2+8=\left(\frac{1\pm(2c-1)}{c-1}+c \right)^2$
makes it slightly easier.  
Taking the $-$ gives us $c=-1 \implies b=-1$ contradicting distinct again, and taking the $+$ gives us the more ugly cubic equation 
$c^3-2c^2+4c-2=0$
By the Rational Root Theorem, one need only test the values $\pm 1, \pm 2$, none of which yield a solution to this cubic, so this cubic has no rational solutions.  
Therefore, the only solution was the first one we found; namely $\boxed{a=1, b=-2,c=0}$.  
