Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-26x+168=0$
$II.$ $y^{2}-25y+156=0$

The condition that ${x^3} - 3px + 2q$ may be divisible by a factor of the form ${x^2} + 2ax + {a^2}$ is

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$,then $\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b} = $

If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 2x + 4 = 0$,then $\frac{1}{\alpha^3} + \frac{1}{\beta^3}$ is equal to

Let $p, q$ and $r$ be real numbers $(p \ne q, r \ne 0),$ such that the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign. Then the sum of squares of these roots is equal to:

The number of integral values of $m$ for which the quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$ is always positive for all $x \in R$ is: