Question

Using the properties of determinants prove that $\left[\begin{array}{ccc}\alpha & {\alpha }^2& \beta +\gamma \\ \beta & {\beta }^2& \gamma +\alpha \\ \gamma & {\gamma }^2& \alpha +\beta \end{array}\right]=(\alpha -\beta )(\beta -\gamma )(\gamma -\alpha )(\alpha +\beta +\gamma )$

Answer 1

$LHS=\left[\begin{array}{ccc}\alpha & {\alpha }^2& \beta +\gamma \\ \beta & {\beta }^2& \gamma +\alpha \\ \gamma & {\gamma }^2& \alpha +\beta \end{array}\right]$

applying $C_1\to C_1+C_3$
$=\left[\begin{array}{ccc}\alpha +\beta +\gamma & {\alpha }^2& \beta +\gamma \\ \beta +\alpha +\gamma & {\beta }^2& \gamma +\alpha \\ \gamma +\alpha +\beta & {\gamma }^2& \alpha +\beta \end{array}\right]$

Now, taking $\alpha +\beta +\gamma$ common from $C_1$
$=(\alpha +\beta +\gamma )\left[\begin{array}{ccc}1& {\alpha }^2& \beta +\gamma \\ 1& {\beta }^2& \gamma +\alpha \\ 1& {\gamma }^2& \alpha +\beta \end{array}\right]$

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we get -
$=(\alpha +\beta +\gamma )\left[\begin{array}{ccc}1& {\alpha }^2& \beta +\gamma \\ 0& (\beta -\alpha )(\beta +\alpha )& -(\beta -\alpha )\\ 0& (\gamma -\alpha )(\gamma +\alpha )& -(\gamma -\alpha )\end{array}\right]$

Now, on taking $(\gamma -\alpha )$ and $(\beta -\alpha )$ common from $R_3$ and $R_2$ resepectively, we get
$=(\alpha +\beta +\gamma )(\beta -\alpha )(\gamma -\alpha )\left[\begin{array}{ccc}1& {\alpha }^2& \beta +\gamma \\ 0& (\beta +\alpha )& -1\\ 0& (\gamma +\alpha )& -1\end{array}\right]$

Now expanding the determinant along $R_1$, we get -
$=(\alpha +\beta +\gamma )(\beta -\alpha )(\gamma -\alpha )(-\beta -\alpha +\gamma +\alpha )=(\alpha +\beta +\gamma )(\beta -\alpha )(\gamma -\alpha )(-\beta +\gamma )=(\alpha -\beta )(\beta -\gamma )(\gamma -\alpha )(\alpha +\beta +\gamma )$

$LHS=RHS$
Hence proved.