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(D) Given equation: $12 x^{1/3} - 25 x^{1/6} + 12 = 0$. Let $t = x^{1/6}$. Then the equation becomes $12 t^2 - 25 t + 12 = 0$. Factoring the quadratic

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$\frac{16}{9}$

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(D) Given equation: $12 x^{1/3} - 25 x^{1/6} + 12 = 0$. Let $t = x^{1/6}$. Then the equation becomes $12 t^2 - 25 t + 12 = 0$. Factoring the quadratic: $12 t^2 - 16 t - 9 t + 12 = 0 \Rightarrow 4 t(3 t - 4) - 3(3 t - 4) = 0$. $(4 t - 3)(3 t - 4) = 0$,so $t = \frac{3}{4}$ or $t = \frac{4}{3}$. Since $\alpha > \beta$ and $x^{1/6} = t$,we have $\alpha^{1/6} = \frac{4}{3}$ and $\beta^{1/6} = \frac{3}{4}$. We need to find $\sqrt[6]{\frac{\alpha}{\beta}} = \frac{\alpha^{1/6}}{\beta^{1/6}}$. Substituting the values: $\frac{4/3}{3/4} = \frac{4}{3} 	imes \frac{4}{3} = \frac{16}{9}$.

$\alpha$ and $\beta$ are the real roots of the equation $12 x^{1/3} - 25 x^{1/6} + 12 = 0$. If $\alpha > \beta$,then $\sqrt[6]{\frac{\alpha}{\beta}} =$

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