Question

Show that:
${\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|}^2=\left|\begin{array}{ccc}2bc-a^2& c^2& b^2\\ c^2& 2ac-b^2& a^2\\ b^2& a^2& 2ab-c^2\end{array}\right|={(a^3+b^3+c^3-3abc)}^2$

Answer 1

$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=-\left|\begin{array}{ccc}a& c& b\\ b& a& c\\ c& b& a\end{array}\right|$

$=\left|\begin{array}{ccc}-a& c& b\\ -b& a& c\\ -c& b& a\end{array}\right|$

So $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|\cdot \left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=\left|\begin{array}{ccc}-a& c& b\\ -b& a& c\\ -c& b& a\end{array}\right|\cdot \left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$

As $detA.detB=det\left(AB\right)$ for two matrices

Here $AB$$=\left[\begin{array}{ccc}-a& c& b\\ -b& a& c\\ -c& b& a\end{array}\right]\cdot \left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$

$=\left[\begin{array}{ccc}-a.a+bc+bc& -ab+c^2+ab& -ac+ac+b^2\\ -ba+ba+c^2& -b^2+ac+ac& -bc+c^2+bc\\ -ca+b^2+ac& -bc+bc+c^2& -c^2+ab+ab\end{array}\right]=\left[\begin{array}{ccc}2bc-a^2& c^2& b^2\\ c^2& 2ac-b^2& a^2\\ b^2& a^2& 2ab-c^2\end{array}\right]$

So |AB|=|A|.|B|$

ie $\left|\begin{array}{ccc}2bc-a^2& c^2& b^2\\ c^2& 2ac-b^2& a^2\\ b^2& a^2& 2ab-c^2\end{array}\right|=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|\left|\begin{array}{ccc}-a& c& b\\ -b& a& c\\ -c& b& a\end{array}\right|$

$\left|\begin{array}{ccc}2bc-a^2& c^2& b^2\\ c^2& 2ac-b^2& a^2\\ b^2& a^2& 2ab-c^2\end{array}\right|=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$

$\left|\begin{array}{ccc}2bc-a^2& c^2& b^2\\ c^2& 2ac-b^2& a^2\\ b^2& a^2& 2ab-c^2\end{array}\right|={\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|}^2$

${\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|}^{}=a(bc-a^2)-b(b^2-ac)+c(ab-c^2)$

${\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|}^{}=3abc-a^3-b^3-c^3$

${\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|}^2={(a^3+b^3+c^3-3abc)}^2$

$\left|\begin{array}{ccc}2bc-a^2& c^2& b^2\\ c^2& 2ac-b^2& a^2\\ b^2& a^2& 2ab-c^2\end{array}\right|={\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|}^2$$={(a^3+b^3+c^3-3abc)}^2$