Let the roots of the equation E_1 equiv x^3+x | vedclass

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(C) Let $x_1, x_2, x_3$ be the roots of the equation $E_1: x^3+x^2+lx+n=0$. From Vieta's formulas,$x_1+x_2+x_3 = -1$. Given $E_2: x^3+ax^2+bx+c=0$ is

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$\sqrt{3}i, -\sqrt{3}i, \frac{-1+\sqrt{3}i}{2}, \frac{-1-\sqrt{3}i}{2}$

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(C) Let $x_1, x_2, x_3$ be the roots of the equation $E_1: x^3+x^2+lx+n=0$. From Vieta's formulas,$x_1+x_2+x_3 = -1$. Given $E_2: x^3+ax^2+bx+c=0$ is a reciprocal equation of class one,so $c=1$ and $a=b$. Thus,$E_2: x^3+ax^2+ax+1=0$. The roots of $E_2$ are $\frac{x_i-1}{2}$. The sum of roots of $E_2$ is $\sum \frac{x_i-1}{2} = \frac{(x_1+x_2+x_3)-3}{2} = \frac{-1-3}{2} = -2$. From $E_2$,the sum of roots is $-a$,so $-a = -2 \Rightarrow a=2$. $E_2$ becomes $x^3+2x^2+2x+1 = (x+1)(x^2+x+1) = 0$. The roots of $E_2$ are $-1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$. Using $\frac{x_i-1}{2} = y_i$,we have $x_i = 2y_i+1$. For $y_1 = -1$,$x_1 = 2(-1)+1 = -1$. For $y_2 = \frac{-1+i\sqrt{3}}{2}$,$x_2 = 2(\frac{-1+i\sqrt{3}}{2})+1 = i\sqrt{3}$. For $y_3 = \frac{-1-i\sqrt{3}}{2}$,$x_3 = 2(\frac{-1-i\sqrt{3}}{2})+1 = -i\sqrt{3}$. The roots of $E_1$ are $\{-1, i\sqrt{3}, -i\sqrt{3}\}$. The roots of $E_2$ are $\{-1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}\}$. The common root is $-1$. Excluding the common root,the remaining roots are $\{i\sqrt{3}, -i\sqrt{3}, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}\}$.

Let the roots of the equation $E_1 \equiv x^3+x^2+lx+n=0$ be $x_i, (i=1, 2, 3)$ and the roots of $E_2 \equiv x^3+ax^2+bx+c=0$ be $\frac{x_i-1}{2}$. If the equation $E_2=0$ is a reciprocal equation of class one,then the roots of these two equations excluding the common roots are

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