Question

Prove the following :
$\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|={(a+b+c)}^3.$

Answer 1

Applying $R_2-R_1$, we get

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

Now applying $R_3+R_2$

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

$=-[a(bc-a^2)-b(b^2-ca)+c(ab-c^2)]$

$=-(3abc-a^3-b^3-c^3)=a^3+b^3+c^3-3abc$