If a, b, c are real numbers such that b+c&c+a&a+b +a&a+b&b+c +b&b+c&c+a =0, then show that either a + b + c = 0 or a = b= c.

Let L.H.S= b+c&c+a&a+b +a&a+b&b+c +b&b+c&c+a = 2(a+b+c)&2(a+b+c)&2(a+b+c) +a&a+b&b+c +b&b+c&c+a [Applying R1 → R1 + R2 + R3] =2(a+b+c) 1&1&1 +a&a+b&b+c +b&b+c&c+a =2(a+b+c) 1&0&0 +a&b-c&b-a +b&c-a&c-b [Applying C2 → C2 - C1 and C3 → C3 - C1] =2(a+b+c) 1 b-c&b-a -a&c-b =2(a+b+c) (b-c)(c-b)-(b-a)(c-a) =-2(a+b+c) a^2+b^2+c^2-ab-bc-ca =-(a+b+c) 2a^2+2b^2+2c^2-2ab-2bc-2ca =-(a+b+c) (a-b)^2+(b-c)^2+(c-a)^2 But =0 [Given] -(a+b+c) (a-b)^2+(b-c)^2+(c-a)^2 =0 Either, (a+b+c)=0 or (a-b)^2+(b-c)^2+(c-a)^2=0 (a+b+c)=0 or a=b=c Hence proved.

If a, b, c are real numbers such that b+c&c+a&a+b +a&a+b&b+c +b&b…

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Question

If a, b, c are real numbers such that $\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}=0,$ then show that either a + b + c = 0 or a = b= c.

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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=\begin{vmatrix}2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$ [Applying R₁ → R₁ + R₂ + R₃]
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}-\text{a}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix}$ [Applying C₂ → C₂ - C₁ and C₃ → C₃ - C₁]
$=2(\text{a}+\text{b}+\text{c})\left\{1\begin{vmatrix}\text{b}-\text{c}&\text{b}-\text{a}\\\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix}\right\}$
$=2(\text{a}+\text{b}+\text{c})\{(\text{b}-\text{c})(\text{c}-\text{b})-(\text{b}-\text{a})(\text{c}-\text{a})\}$
$=-2(\text{a}+\text{b}+\text{c})\{\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}$
But $\triangle=0$ [Given]
$\Rightarrow-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}=0$
Either,
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2=0$
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $\text{a}=\text{b}=\text{c}$
Hence proved.

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