Let E_1 equiv ax^2+bx+c,E_2 equiv bx^2+cx+a,E | vedclass

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(D) Given $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3$,we multiply by $abc$ to get $a^3+b^3+c^3=3abc$. This implies either $a+b+c=0$ or $a=b=c$. I

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$x^2-\frac{a(b+c)x}{bc}+\frac{a^2}{bc}$

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(D) Given $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3$,we multiply by $abc$ to get $a^3+b^3+c^3=3abc$. This implies either $a+b+c=0$ or $a=b=c$. If $a=b=c$,then $E_1=E_2=E_3=a(x^2+x+1)$,which have the same roots. If $a+b+c=0$,then $x=1$ is a root for $E_1, E_2, E_3$ because $a(1)^2+b(1)+c = a+b+c=0$,$b(1)^2+c(1)+a = b+c+a=0$,and $c(1)^2+b(1)+a = c+b+a=0$. For $E_2$ and $E_3$,the roots are $x=1$ and $x=\frac{a}{b}$ (from $E_2$) and $x=\frac{a}{c}$ (from $E_3$). The common root is $x=1$. The other roots are $x=\frac{a}{b}$ and $x=\frac{a}{c}$. The quadratic expression with roots $\frac{a}{b}$ and $\frac{a}{c}$ is $(x-\frac{a}{b})(x-\frac{a}{c}) = x^2 - (\frac{a}{b}+\frac{a}{c})x + \frac{a^2}{bc} = x^2 - \frac{a(b+c)}{bc}x + \frac{a^2}{bc}$.

Let $E_1 \equiv ax^2+bx+c$,$E_2 \equiv bx^2+cx+a$,$E_3 \equiv cx^2+bx+a$ and $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3$. If these quadratic expressions have a common zero,then the quadratic expression having zeroes that are common to $E_2$ and $E_3$ and different from the zeroes of $E_1$ is

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