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(C) Expanding the given equation: $(x^2 - (b+c)x + bc) + (x^2 - (a+c)x + ac) + (x^2 - (a+b)x + ab) = 0$ $3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0$ The discri

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(C) Expanding the given equation: $(x^2 - (b+c)x + bc) + (x^2 - (a+c)x + ac) + (x^2 - (a+b)x + ab) = 0$ $3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0$ The discriminant $D$ of this quadratic equation is given by: $D = [-2(a+b+c)]^2 - 4(3)(ab+bc+ca)$ $D = 4(a+b+c)^2 - 12(ab+bc+ca)$ $D = 4(a^2+b^2+c^2 + 2ab+2bc+2ca - 3ab-3bc-3ca)$ $D = 4(a^2+b^2+c^2 - ab-bc-ca)$ $D = 2[(a-b)^2 + (b-c)^2 + (c-a)^2]$ Since $a, b, c$ are real,$(a-b)^2, (b-c)^2, (c-a)^2 \geq 0$. Therefore,$D \geq 0$. Since the discriminant is always non-negative,the roots are always real.

If $a, b, c$ are real numbers,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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