The minimum value of the sum of the squares of the roots of $$x^{2}+(3-a) x+1=2 a$$ is:

If $$\alpha, \beta, \gamma, \delta$$ are the roots of the equation $$x^{4}+x^{3}+x^{2}+x+1=0$$, then $$\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}$$ is equal to :

Let $${S_1} = \left\{ {x \in R - \{ 1,2\} :{{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0} \right\}$$ and $${S_2} = \left\{ {x \in R:{3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0} \right\}$$. Then, $${S_1} \cup {S_2}$$ is equal to :

Let S be the set of all integral values of $$\alpha$$ for which the sum of squares of two real roots of the quadratic equation $$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$$ is minimum. Then S :