If a + b + c ¹ 0 and then prove that a = b = c

Solution:

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

$\left[\right.$ Applying $\left.R_{1} \rightarrow R_{1}+R_{2}+R_{3}\right]$

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

Now,

[Applying $C_{1} \rightarrow C_{1}-C_{3}$ and $C_{2} \rightarrow C_{2}-C_{3}$ ]

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

[Expanding along $R_{1}$ ]

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

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

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

$=\frac{-1}{2}(a+b+c)\left[2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a\right]$

$=-\frac{1}{2}(a+b+c)\left[\left(a^{2}+b^{2}-2 a b\right)+\left(b^{2}+c^{2}-2 b c\right)+\left(c^{2}+a^{2}-2 a c\right)\right]$

$=\frac{-1}{2}(a+\dot{b}+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$

Given, $\Delta=0$

$\frac{-1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0$

$(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0 \quad[\because a+b+c \neq 0$, given $]$

$a-b \pm b-c=c-a=0$

$a=b=c$