Question

If the equation
$x+ay+(a^2-bc)=0$
$x+by+(b^2-ca)=0$
$x+cy+(c^2-ab)=0$
Consider: $\left[\begin{array}{ccc}1& a& a^2-bc\\ 1& b& b^2-ca\\ 1& c& c^2-ab\end{array}\right]=0$

Answer 1

$\begin{array}{c}\Rightarrow \left(\begin{array}{ccc}1& a& a^2-bc\\ 1& b& b^2-ca\\ 1& c& c^2-ab\end{array}\right)\\ \Rightarrow R_2\to R_2-R_{1}and{R}_3\to R_3-R_1\\ \Rightarrow \left(\begin{array}{ccc}1& a& a^2-bc\\ 0& \left(b-a\right)& \left(b^2-ca-a^2+bc\right)\\ 0& \left(c-a\right)& \left(c^2-ab-a^2+bc\right)\end{array}\right)\\ onexpanding{C}_1\\ \Rightarrow \left(\begin{array}{cc}\left(b-a\right)& \left(b^2-ca-a^2+bc\right)\\ \left(c-a\right)& \left(c^2-ab-a^2+bc\right)\end{array}\right)=\left(\begin{array}{cc}\left(b-a\right)& \left(b-a\right)\left(b+a\right)c\left(b-a\right)\\ \left(c-a\right)& \left(c-a\right)\left(c+a\right)+b\left(c-a\right)\end{array}\right)\\ =0proved.\end{array}$