Question

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

Answer 1

Lets take the LHS of the given expression
$\Delta =\left|\begin{array}{ccc}a+b+2c& a& b\\ c& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|$
Perform $C_1\to C_1+C_2+C_3$

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

Now, take $2(a+b+c)$ common from $C_1$
$=2(a+b+c)\left|\begin{array}{ccc}1& a& b\\ 1& b+c+2a& b\\ 1& a& c+a+2b\end{array}\right|$

Perform $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$

$=2(a+b+c)\left|\begin{array}{ccc}0& -(a+b+c)& 0\\ 0& b+c+a& -(c+a+b)\\ 1& a& c+a+2b\end{array}\right|$

Now, take $(a+b+c)$ common from $R_1$ and $R_2$
$=2(a+b+c{)}^3\left|\begin{array}{ccc}0& -1& 0\\ 0& 1& -1\\ 1& a& c+a+2b\end{array}\right|$

Now, expand the determinants,
$=2(a+b+c{)}^3\times 1(1-0)$
$=2(a+b+c{)}^3$ $Proved$