Question

If $\left|\begin{array}{ccc}3& a+b+c& a^2+b^2+c^2\\ a+b+c& a^2+b^2+c^2& a^3+b^3+c^3\\ a^2+b^2+c^2& a^3+b^3+c^3& a^4+b^4+c^4\end{array}\right|={\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ a& b& c\\ a^2& b^2& c^2\end{array}\right|}^2,$ then the value of $\alpha +\beta +\gamma$ is

Answer 1

$\left|\begin{array}{ccc}1+1+1& a+b+c& a^2+b^2+c^2\\ a+b+c& a^2+b^2+c^2& a^3+b^3+c^3\\ a^2+b^2+c^2& a^3+b^3+c^3& a^4+b^4+c^4\end{array}\right|$
By observing, determinant can be split into the mutiple of two determinants, we get
$=\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$
$\Rightarrow \alpha +\beta +\gamma =1+1+1=3$