Question

Let $\alpha, \beta, \gamma$ are the real roots of the equation $x^{3}+a x^{2}+b x+c=0(a, b, c \in R$ and $a \neq 0)$
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

Roots of equation $x^{3}+a x^{2}+b x+c=0$ 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
$\begin{vmatrix}\alpha&\beta&\gamma\\ \beta&\gamma&\alpha\\ \gamma&\alpha&\beta\end{vmatrix}=0$
$\Rightarrow \alpha^{3}+\beta^{3}+\gamma^{3}-3 \alpha \beta \gamma=0$
$\Rightarrow (\alpha+\beta+\gamma)\left[\alpha^{2}+\beta^{2}+\gamma^{2}-\alpha \beta-\beta \gamma-\gamma \alpha\right]=0$
$\Rightarrow (\alpha+\beta+\gamma)\left[(\alpha+\beta+\gamma)^{2}-3(\alpha \beta+\beta \gamma+\gamma \alpha)\right]=0$
$\Rightarrow -a\left[a^{2}-3 b\right]=0$
Hence $a^{2}=3 b$ (as $a \neq 0)$
$\Rightarrow a^{2} / b=3$