Suppose $\alpha, \beta, \gamma$ are roots of $x^3+x^2+2x+3=0$. If $f(x)=0$ is a cubic polynomial equation whose roots are $\alpha+\beta, \beta+\gamma, \gamma+\alpha$,then $f(x)$ is equal to

If $a, b$ and $c$ are the roots of $x^3+qx+r=0$,then $(a-b)^2+(b-c)^2+(c-a)^2=$ (in $q$)

The difference between two roots of the equation $x^3-13x^2+15x+189=0$ is $2$. Then the roots of the equation are:

If $\alpha$ and $\beta$ are the roots of the equation $x^2-ax+b=0$,and $\alpha^2+\beta^2$ and $\alpha^3+\beta^3$ are the roots of the equation $Ax^2+Bx+C=0$,then $C=$

Suppose that the equation $ax^2+bx+c=0$ has roots $\alpha$ and $\beta$,both of which are different from $\frac{1}{3}$. Then,an equation whose roots are $\frac{1}{3\alpha-1}$ and $\frac{1}{3\beta-1}$ is

If $\alpha, \beta$ are the roots of the equation $x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0$,then find the equation whose roots are $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$.