Let a, b, c be non-zero real roots of the equ | vedclass

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(C) Given the cubic equation $x^3+ax^2+bx+c=0$ with roots $a, b, c$.By Vieta's formulas:$a+b+c = -a \implies 2a+b+c = 0$ $(i)$$ab+bc+ca = b$ (ii)$abc

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There are exactly two such triples $(a, b, c)$

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(C) Given the cubic equation $x^3+ax^2+bx+c=0$ with roots $a, b, c$.By Vieta's formulas:$a+b+c = -a \implies 2a+b+c = 0$ $(i)$$ab+bc+ca = b$ (ii)$abc = -c \implies ab = -1$ (since $c 
eq 0$) (iii)From (iii),$b = -\frac{1}{a}$.Substitute $b$ into $(i)$: $2a - \frac{1}{a} + c = 0 \implies c = \frac{1}{a} - 2a$.Substitute $b$ and $c$ into (ii):$ab + c(a+b) = b$$-1 + (\frac{1}{a} - 2a)(a - \frac{1}{a}) = -\frac{1}{a}$$-1 + (1 - \frac{1}{a^2} - 2a^2 + 2) = -\frac{1}{a}$$2 - \frac{1}{a^2} - 2a^2 = -\frac{1}{a}$Multiply by $a^2$: $2a^2 - 1 - 2a^4 = -a$$2a^4 - 2a^2 - a + 1 = 0$$(a-1)(2a^3+2a^2-1) = 0$.For $a=1$,$b=-1$,$c=-1$. Roots are $1, -1, -1$. Check: $x^3+x^2-x-1=0 \implies (x-1)(x+1)^2=0$. Roots are $1, -1, -1$. This works.For $2a^3+2a^2-1=0$,there is one real root $a \approx 0.589$. This yields a second valid triple $(a, b, c)$.Thus,there are exactly two such triples.

Let $a, b, c$ be non-zero real roots of the equation $x^3+ax^2+bx+c=0$. Then,

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