Question

If $\left|\begin{array}{ccc}(b+c{)}^2& a^2& a^2\\ b^2& (a+c{)}^2& b^2\\ c^2& c^2& (a+b{)}^2\end{array}\right|=kabc(a+b+c{)}^3,$
then $k=$

Answer 1

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

$C_1\to C_1-C_2,$
$=\left|\begin{array}{ccc}(b+c-a)(b+c+a)& a^2& a^2\\ (b-a-c)(b+a+c)& (a+c{)}^2& b^2\\ 0& c^2& (a+b{)}^2\end{array}\right|$

$C_2\to C_2-C_3,$
$=\left|\begin{array}{ccc}(b+c-a)(b+c+a)& 0& a^2\\ (b-a-c)(b+a+c)& (a+c-b)(a+c+b)& b^2\\ 0& (c-a-b)(c+a+b)& (a+b{)}^2\end{array}\right|$

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

$R_2\to R_2-R_1,R_2\to R_2-R_3$
$=2(a+b+c{)}^2\left|\begin{array}{ccc}(b+c-a)& 0& a^2\\ -c& a& -a(a+b)\\ 0& (c-a-b)& (a+b{)}^2\end{array}\right|$
Expanding along $R_1$, we get
$=2abc(a+b+c{)}^3$

$\Rightarrow k=2$

Alternate Solution:
Put $a=b=c=1$
$\left|\begin{array}{ccc}4& 1& 1\\ 1& 4& 1\\ 1& 1& 4\end{array}\right|=27k\Rightarrow 54=27k\Rightarrow k=2$