If the difference between the roots of the equation $x^2 + px + q = 0$ is $1$,then:

Let $p$ and $q$ be real numbers such that $p \neq 0, p^3 \neq q$ and $p^3 \neq -q$. If $\alpha$ and $\beta$ are non-zero numbers satisfying $\alpha + \beta = -p$ and $\alpha^3 + \beta^3 = q$,then find the quadratic equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.

Let $\tan 30^{\circ}$ and $\tan 15^{\circ}$ be the roots of the quadratic equation $x^2+ax+b=0$,then $1+a-b=$

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 $q^2 - p^2$ is equal to:

If $\alpha$ and $\beta$ are two roots of the equation $x^{2}-64x+256=0$,then the value of $\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{\frac{1}{8}}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{\frac{1}{8}}$ is

If $p$ and $q$ are the roots of the equation $x^{2}+px+q=0$, then: