Solve the following
Question:

Let $\alpha, \beta, \gamma$ be the real roots of the equation, $x^{3}+a x^{2}+b x+c=0,(a, b, c \in R$ and $a, b \neq 0)$ If the system of equations (in, $\mathrm{u}, \mathrm{v}, \mathrm{w}$ ) given by $\alpha \mathrm{u}+\beta \mathrm{v}+\gamma_{\mathrm{w}}=0, \beta \mathrm{u}+\gamma \mathrm{v}+\alpha \mathrm{w}=0$ $\gamma_{\mathrm{u}}+\alpha_{\mathrm{v}}+\beta_{\mathrm{w}}=0$ has non-trivial solution, then the value of $\frac{\mathrm{a}^{2}}{\mathrm{~b}}$ is

1. (1) 5

2. (2) 3

3. (3) 1

4. (4) 0

Correct Option: , 2

Solution:

$\left|\begin{array}{lll}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta\end{array}\right|=0$

$\Rightarrow-(\alpha+\beta+\gamma)\left(\alpha^{2}+\beta^{2}+\gamma^{2}-\sum \alpha \beta\right)=0$

$\Rightarrow-(-a)\left(a^{2}-2 b-b\right)=0$

$\Rightarrow a\left(a^{2}-3 b\right)=0$

$\Rightarrow a^{2}=3 b \Rightarrow \frac{a^{2}}{b}=3$