If a, b and c are real numbers, and = b+c&c+a&a+b +a&a+b&b+c +b&b+c&c+a =0, Show that either a + b + c = 0 or a = b = c.

= b+c&c+a&a+b +a&a+b&b+c +b&b+c&c+a Applying R_1 R_1 + R_2 + R_3, we have: = 2(a+b+c)&2(a+b+c)&2(a+b+c) +a&a+b&b+c +b&b+c&c+a =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&a-b&b-a +b&c-a&c-b C_1 _2-C_1 and C_3 _3-C_1 Expanding along R_1, we have: = 2(a + b + c) (1) [(b - c) (c - b) - (b - a) (c - a)] = 2(a + b + c) [-b^2- c^2 + 2bc - bc + ba + ac - a^2] = 2(a + b + c) [ab + bc + ca - a^2 - b^2 - c^2] It is given that = 0. (a + b + c) [ab + bc + ca - a^2 - b^2 - c^2] = 0 Either a + b + c = 0, or ab + bc + ca - a^2 - b^2 - c^2 = 0. Now, ab + bc + ca - a^2 - b^2 - c^2 = 0 -2ab - 2bc - 2ca + 2a^2 + 2b^2 + 2c^2 = 0 (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 (a -b)^2= (b - c)^2 = (c - a)^2 = 0 [(a - b)^2, (b - c)^2, (c - a)^2 are non-negative] (a - b) = (b - c) = (c - a) = 0 a = b = c Hence, if = 0, then either a + b + c = 0 or a = b = c.

If a, b and c are real numbers, and = b+c&c+a&a+b +a&a+b&b+c +b&b…

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If a, b and c are real numbers, and
$\triangle=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}=0,$
Show that either a + b + c = 0 or a = b = c.

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Answer

$\triangle=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
Applying $R_1\rightarrow R_1 + R_2 + R_3, $we have:
$\triangle=\begin{vmatrix}2(a+b+c)&2(a+b+c)&2(a+b+c)\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$=2(a+b+c)\begin{vmatrix}1&1&1\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$=2(a+b+c)\begin{vmatrix}1&0&0\\c+a&a-b&b-a\\a+b&c-a&c-b\end{vmatrix}\ \text{C}_1\rightarrow\text{C}_2-\text{C}_1\ \text{and}\ \text{C}_3\rightarrow\text{C}_3-\text{C}_1$
Expanding along $R_1$, we have:
$\triangle$ $= 2(a + b + c) (1) [(b - c) (c - b) - (b - a) (c - a)]$
$= 2(a + b + c) [-b^2- c^2 + 2bc - bc + ba + ac - a^2]$
$= 2(a + b + c) [ab + bc + ca - a^2 - b^2 - c^2]$
It is given that $\triangle$
$= 0.$
$(a + b + c) [ab + bc + ca - a^2 - b^2 - c^2] = 0$
$\Rightarrow $ Either $a + b + c = 0, or ab + bc + ca - a^2 - b^2 - c^2 = 0.$
Now,
$ab + bc + ca - a^2 - b^2 - c^2 = 0$
$\Rightarrow -2ab - 2bc - 2ca + 2a^2 + 2b^2 + 2c^2 = 0$
$\Rightarrow (a - b)^2 + (b - c)^2 + (c - a)^2 = 0$
$\Rightarrow (a -b)^2= (b - c)^2 = (c - a)^2 = 0 [(a - b)^2, (b - c)^2, (c - a)^2$ are non-negative]
$\Rightarrow (a - b) = (b - c) = (c - a) = 0$
$\Rightarrow a = b = c$
Hence, if $\triangle$ = 0, then either$ a + b + c = 0 or a = b = c.$

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