If $p$ and $q$ are non-zero real numbers and $\alpha^3 + \beta^3 = -p$,$\alpha \beta = q$,then a quadratic equation whose roots are $\frac{\alpha^2}{\beta}$ and $\frac{\beta^2}{\alpha}$ is

The sum of the roots of a quadratic equation is $2$ and the sum of their cubes is $98$. Then the equation is:

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2}+bx+c=0$ and $3b^{2}=16ac$, then:

If $\alpha, \beta$ are the roots of $x^2 + px + 1 = 0$ and $\gamma, \delta$ are the roots of $x^2 + qx + 1 = 0$,then $(\alpha - \gamma) (\beta - \gamma) (\alpha + \delta) (\beta + \delta) = \dots$

Let $\alpha, \beta$ be the roots of the equation $x^2-ax-b=0$ with $\operatorname{Im}(\alpha) < \operatorname{Im}(\beta)$. Let $P_n=\alpha^n-\beta^n$. If $P_3=-5 \sqrt{7} i, P_4=-3 \sqrt{7} i, P_5=11 \sqrt{7} i$ and $P_6=45 \sqrt{7} i$,then $|\alpha^4+\beta^4|$ is equal to . . . . . .

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + x^2 + x + 1 = 0$,then match the items of List-$I$ with those of List-$II$:
List-$I$:
$(i)$ $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$
(ii) $\alpha^3 + \beta^3 + \gamma^3$
(iii) $\alpha^4 + \beta^4 + \gamma^4$
(iv) $(\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2$
List-$II$:
$(A)$ $-1$
$(B)$ $-4$
$(C)$ $1$
$(D)$ $3$
$(E)$ $0$