જો $a, b$ અને $c$ વાસ્તવિક સંખ્યાઓ હોય, અને $\Delta=\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|=0$ હોય, તો સાબિત કરો કે $a+b+c=0$ અથવા $a=b=c$.

$\Delta=\left|\begin{array}{lll}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|=0$

Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we have:

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

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

Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1},$ we have:

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

Expanding along $R_{1},$ we have:

$\Delta=2(a+b+c)(1)[(b-c)(c-b)-(b-a)(c-a)]$

$=2(a+b+c)\left[-b^{2}-c^{2}+2 b c-b c+b a+a c-a^{2}\right]$

$=2(a+b+c)\left[a b+b c+c a-a^{2}-b^{2}-c^{2}\right]$

It is given that $\Delta=0$ $(a+b+c)\left[a b+b c+c a-a^{2}-b^{2}-c^{2}\right]=0$

$\Rightarrow$ Either $a+b+c=0,$ or $a b+b c+c a-a^{2}-b^{2}-c^{2}=0$

Now, $a b+b c+c a-a^{2}-b^{2}-c^{2}=0$

$\Rightarrow-2 a b-2 b c-2 c a+2 a^{3}+2 b^{3}+2 c^{3}=0$

$\Rightarrow(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$

$\Rightarrow(a-b)=(b-c)^{2}=(c-a)^{2}=0 \quad\left[(a-b)^{2},(b-c)^{2},(c-a)^{2} \text { are non-negative }\right]$

$\Rightarrow(a-b)=(b-c)=(c-a)=0$

$\Rightarrow a=b=c$

Hence, if $\Delta=0,$ then either $a+b+c=0$ or $a=b=c$