Let $\alpha$, $\beta$, $\gamma$ be the real roots of the equation, x³ + ax² + bx + c = 0, (a, b, c $\in$ R and a, b $\ne$ 0). If the system of equations (in u, v, 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 solution, then the value of ${{{a^2}} \over b}$ is

x³ + ax² + bx + c = 0 <br><br>Roots are $\alpha$, $\beta$, $\gamma$.<br><br>For non-trivial solutions,<br><br>$$\left| {\matrix{ \alpha &amp; \beta &amp; \gamma \cr \beta &amp; \gamma &amp; \alpha \cr \gamma &amp; \alpha &amp; \beta \cr } } \right| = 0$$<br><br>$\Rightarrow$ ${\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = 0$<br><br>$\Rightarrow$ $$(\alpha + \beta + \gamma )\left[ {{{\left( {\alpha + \beta + \alpha } \right)}^2} - 3\left( {\sum {\alpha \beta } } \right)} \right] = 0$$<br><br>$\Rightarrow$ $( - a)[{a^2} - 3b] = 0$<br><br>$\Rightarrow$ ${a^2} = 3b$ ($\because$ a $\ne$ 0)<br><br>$\Rightarrow {{{a^2}} \over b} = 3$