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If $a, b, c$ are real, then both the roots of the equation $(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$ are always
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The correct answer is:
real
Hints : $3 x^2-2 x(a+b+c)+a b+b c+c a=0$
$$
\begin{aligned}
& \mathrm{D}=4(a+b+c)^2-4.3(a b+b c+c a) \\
& =4\left(a^2+b^2+c^2-a b-b c-c a\right) \\
& =2\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\
& =\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\
& \geq 0
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{D}=4(a+b+c)^2-4.3(a b+b c+c a) \\
& =4\left(a^2+b^2+c^2-a b-b c-c a\right) \\
& =2\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\
& =\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\
& \geq 0
\end{aligned}
$$
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