Finding the rational values of constant for which these constants are roots of equation Problem : 
Determine all rational values for which $a,b,c$ are the roots of $x^3+ax^2+bx+c=0$
Solution : 
Sum of the roots $a+b+c = -a$ ........(i) ( Since , as per question $a,b,c$ are roots of equation and we have to find values ) 
$\sum_{a,b,c} ab = b ........(ii)$ 
$abc= -c \Rightarrow ab = -1 \Rightarrow a =\frac{-1}{b} .....(iii)$
Solving (i),(ii) and (iii) simultaneously we get an equation which is biquadratic in "a" which is : 
$2a^4-2a^2-a+1=0$ Here we can find that $a = 1$ is a factor of this equation and by long division of polynomial we can divide $a-1$ with the given biquadratic we get cubic polynomial in a which is $2a^3+2a^2-1$ ...
Please guide further and suggest whether this is correct or not.. Thanks...

Neither $-1$ nor $1$ is a solution to your final cubic, so this calls for the use of the cubic formula.
$$a = 2, b = 2, c = 0, d = -1$$
$$\Delta_0 = 2^2 - 3(2)(0) = 4$$
$$\Delta_1 = 2(2^3) - 9(2)(2)(0) + 27(2^2)(-1) = 16 - 108 = -92$$
$$C = \sqrt[3]{\frac{-92 + \sqrt{(-92)^2 - 4(4^3)}}{2}} = \sqrt[3]{\frac{-92 + \sqrt{8208}}{2}} = \sqrt[3]{\frac{12\sqrt{57} - 92}{2}} = \sqrt[3]{6\sqrt{57} - 46}$$
We can divide $8208$ by $-27a^2$ (or $-108$) to get the equation's discriminant, $-76$. This means the equation has one real root and two complex roots. That real root is given by:
$$-\frac{1}{6}\left(2 + \sqrt[3]{6\sqrt{57} - 46} + \frac{4}{\sqrt[3]{6\sqrt{57} - 46}}\right)$$
A number which is definitely not rational. So you have found the only rational root, 1.
