Question

Let $\alpha ,\beta ,\gamma$ are the real roots of the equation $x^3+a{x}^2+bx+c=0(a,b,c\in R$ and $a\ne 0)$
If the system of equations (in $u,v$ and $w$ ) given by
$\alpha u+\beta v+\gamma w=0$
$\beta u+\gamma v+\alpha w=0$
$\gamma u+\alpha v+\beta w=0$
has non-trivial solutions, then the value of $a^2/b$ is equal to_____.

Answer 1

Answer: 3

Roots of equation $x^3+a{x}^2+bx+c=0$ are $\alpha +\beta +\gamma$
$\therefore \alpha +\beta +\gamma =-a$
$\alpha \beta +\beta \gamma +\gamma \alpha =b$
Since the given system of equations has non-trivial solutions, so
$\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|=0$
$\Rightarrow {\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma =0$
$\Rightarrow (\alpha +\beta +\gamma )\left[{\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\gamma \alpha \right]=0$
$\Rightarrow (\alpha +\beta +\gamma )\left[(\alpha +\beta +\gamma {)}^2-3(\alpha \beta +\beta \gamma +\gamma \alpha )\right]=0$
$\Rightarrow -a\left[a^2-3b\right]=0$
Hence $a^2=3b$ (as $a\ne 0)$
$\Rightarrow a^2/b=3$