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If $a, b$ and $c$ are real numbers then the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x$$-a)=0$ are always
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Given equation is $(x-a)(x-b)+(x-b)$ $(x-c)+(x-c)(x-a)=0$
$\Rightarrow 3 x^{2}-2(b+a+c) x+a b+b c+c a=0$
Now, here $A=3, B=-2(a+b+c)$
$C=a b+b c+c a$
$\therefore D=\sqrt{B^{2}-4 A C}$
$=\sqrt{(-2(a+b+c))^{2}-4(3)(a b+b c+c a)}$
$=\sqrt{4(a+b+c)^{2}-12(a b+b c+c a)}$
$=2 \sqrt{a^{2}+b^{2}+c^{2}-a b-b c-c a}$
$=2 \sqrt{\frac{1}{2}\left\{(a-b)^{2}-(b-c)^{2}+(c-a)^{2}\right\}} \geq 0$
$\Rightarrow 3 x^{2}-2(b+a+c) x+a b+b c+c a=0$
Now, here $A=3, B=-2(a+b+c)$
$C=a b+b c+c a$
$\therefore D=\sqrt{B^{2}-4 A C}$
$=\sqrt{(-2(a+b+c))^{2}-4(3)(a b+b c+c a)}$
$=\sqrt{4(a+b+c)^{2}-12(a b+b c+c a)}$
$=2 \sqrt{a^{2}+b^{2}+c^{2}-a b-b c-c a}$
$=2 \sqrt{\frac{1}{2}\left\{(a-b)^{2}-(b-c)^{2}+(c-a)^{2}\right\}} \geq 0$
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