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 $\boxed{a=1, b=-2,c=0}$.
Try sum of roots, product of roots of cubic equation
$\alpha + \beta + \gamma = -\frac ba $
$\alpha\beta + \beta\gamma + \alpha\gamma = \frac ca $
$\alpha \beta \gamma = -\frac da $
The idea provided by @JasonM is correct. You should be able to solve the problem as follows.
Notice that the cubic polynomial is monic, meaning that the coefficient of $x^3$ is $1.$ This is nice because now we can write the polynomial as $(x - a)(x - b)(x - c)$ in factored form. Now, we can equate this to the given form: $$(x - a)(x - b)(x - c) = x^3 + ax^2 + bx + c$$ $$x^3 - (a + b + c)x^2 + (ab + bc + ca)x - abc = x^3 + ax^2 + bx + c.$$
We can see from the equation that $-abc = c,$ meaning that $ab = -1.$ We also have that $a + b + c = a,$ meaning that $b + c = 0,$ or $b = -c.$ With $ab + bc + ca = b,$ substituting values in, we get $-1 - c^2 + 1 = - c,$ which can be rearranged to $c^2 - c = c(c - 1) = 0.$ The solution $c = 0$ cannot be accepted because in that case, $b = c = 0$ and the roots are not distinct. But if $c = 1,$ then $b = -1$ and $a = 1,$ in which case $a = c.$ So there is $\boxed{\text{No solution}}$ to the requested problem.
