Question

Using properties of determinants prove the following questions.

$\left|\begin{array}{ccc}\alpha & {\alpha }^2& \beta +\gamma \\ \beta & {\beta }^2& \gamma +\alpha \\ \gamma & {\gamma }^2& \alpha +\beta \end{array}\right|=(\beta -\gamma )(\gamma -\alpha )(\alpha -\beta )(\alpha +\beta +\gamma )$

Answer 1

Firstly operate $C_3\to C_3+C_1$ and take common $(\alpha +\beta +\gamma )$ from $C_3$ and then easy to expand the determinant.

Let $A=\left|\begin{array}{ccc}\alpha & {\alpha }^2& \beta +\gamma \\ \beta & {\beta }^2& \gamma +\alpha \\ \gamma & {\gamma }^2& \alpha +\beta \end{array}\right|=\left|\begin{array}{ccc}\alpha & {\alpha }^2& \alpha +\beta +\gamma \\ \beta & {\beta }^2& \alpha +\gamma +\beta \\ \gamma & {\gamma }^2& \alpha +\beta +\gamma \end{array}\right|$
(using $C_3\to C_3+C_1$)
$=(\alpha +\beta +\gamma )\left|\begin{array}{ccc}\alpha & {\alpha }^2& 1\\ \beta & \beta & 1\\ \gamma & {\gamma }^2& 1\end{array}\right|$
(Taking out $(\alpha +\beta +\gamma )$ common from $C_1$)
$=(\alpha +\beta +\gamma )\left|\begin{array}{ccc}\alpha & {\alpha }^2& 1\\ \beta -\alpha & {\beta }^2-{\alpha }^2& 0\\ \gamma -\alpha & {\gamma }^2-{\alpha }^2& 0\end{array}\right|$
(using $R_2\to R_2-R_{1}and{R}_3\to R_3-R_1$)
Expanding along $C_3$, we get
$=(\alpha +\beta +\gamma )\left[\right(\beta -\alpha \left)\right({\gamma }^2-{\alpha }^2)-(\gamma -\alpha \left)\right({\beta }^2-{\alpha }^2\left)\right]=(\alpha +\beta +\gamma )\left[\right(\beta -\alpha \left)\right(\gamma -\alpha \left)\right(\gamma +\alpha )-(\gamma -\alpha \left)\right(\beta -\alpha \left)\right(\beta +\alpha \left)\right]=(\alpha +\beta +\gamma )\left[\right(\beta -\alpha \left)\right(\gamma -\alpha \left)\right[\gamma +\alpha -\beta -\alpha \left]\right]=(\alpha +\beta +\gamma )(\alpha -\beta )(\beta -\gamma )(\gamma -\alpha )$
Hence proved