If $x = 2 + 2^{2/3} + 2^{1/3},$ then $x^3 - 6x^2 + 6x = $

Let $S$ be the set of all non-zero real numbers $\alpha$ such that the quadratic equation $\alpha x^2 - x + \alpha = 0$ has two distinct real roots $x_1$ and $x_2$ satisfying the inequality $|x_1 - x_2| < 1$. Which of the following intervals is(are) a subset$(s)$ of $S$?
$(A) \left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right)$
$(B) \left(-\frac{1}{\sqrt{5}}, 0\right)$
$(C) \left(0, \frac{1}{\sqrt{5}}\right)$
$(D) \left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$

If $\alpha$ and $\beta$ are the real roots of the equation $\sqrt{\frac{5x}{x-2}} + \sqrt{\frac{x-2}{5x}} = \frac{29}{10}$ and $\alpha > \beta$,then $\sqrt{\alpha^2 - 11^4 \beta^2} = $

If $\alpha$ is a repeated root of multiplicity $2$ of the equation $18x^3-33x^2+20x-4=0$,then

Find the set $\{ x \in R : |x - 2| = x^2 \}$.

If the equation $x^2 + \lambda x + \mu = 0$ has equal roots and one root of the equation $x^2 + \lambda x - 12 = 0$ is $2$,then $(\lambda, \mu) = $