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(D) Given $x^3-a^3=0 \Rightarrow (x-a)(x^2+ax+a^2)=0$. Since $a>0$,the real root is $\alpha = a$. The other roots are $\beta = a \omega = a(-\frac{1}{

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$1$

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(D) Given $x^3-a^3=0 \Rightarrow (x-a)(x^2+ax+a^2)=0$. Since $a>0$,the real root is $\alpha = a$. The other roots are $\beta = a \omega = a(-\frac{1}{2} + i\frac{\sqrt{3}}{2})$ and $\gamma = a \omega^2 = a(-\frac{1}{2} - i\frac{\sqrt{3}}{2})$. The first curve is $|z - \beta| = \frac{\sqrt{3}a}{2}$,which represents a circle with center $\beta = (-\frac{a}{2}, \frac{a\sqrt{3}}{2})$ and radius $R = \frac{\sqrt{3}a}{2}$. The second curve is $|z - \gamma| = \frac{\sqrt{3}a}{2}$,which represents a circle with center $\gamma = (-\frac{a}{2}, -\frac{a\sqrt{3}}{2})$ and radius $R = \frac{\sqrt{3}a}{2}$. The distance between the centers is $d = \sqrt{(-\frac{a}{2} - (-\frac{a}{2}))^2 + (\frac{a\sqrt{3}}{2} - (-\frac{a\sqrt{3}}{2}))^2} = \sqrt{0 + (a\sqrt{3})^2} = a\sqrt{3}$. The sum of the radii is $R_1 + R_2 = \frac{\sqrt{3}a}{2} + \frac{\sqrt{3}a}{2} = a\sqrt{3}$. Since the distance between the centers $d$ is equal to the sum of the radii $R_1 + R_2$,the two circles touch each other externally at exactly one point. Thus,the number of common points is $1$.

$\alpha$ is the real root and $\beta, \gamma$ are the other roots of the equation $x^3-a^3=0$ $(a>0)$. Then the number of common points of the curves given by $|z-\beta|=\frac{\sqrt{3} a}{2}$ and $|z-\gamma|=\frac{\sqrt{3} a}{2}$ is

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