Cubic equation's determinant

We shall first simplify the determinant and then apply Vieta's Formulas to attain the answer

Let $\Delta = \begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \\ \end{vmatrix}$

Using the elementary transformations

• $R_1 \to R_1 + R_2 + R_3$

• $C_1 \to C_1 - C_2$

• $C_2 \to C_2 - C_3$ ,

and expanding through $R_1$

we get

$\displaystyle \Delta = (\sum\alpha) \times (\sum\alpha\beta - \sum \alpha^2)$

$\displaystyle \Delta = (\sum\alpha) \times (3\sum\alpha\beta - (\sum\alpha)^2)$

Using Vieta's Formulas:

$\displaystyle \sum \alpha = -a$

$\displaystyle \sum \alpha\beta = 0$

Hence:

$\Delta = -a \times - a^2 = \color{#D61F06}{\boxed{a^3}}$