If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3+\ldots$ to $12$ terms $=$

$p$ and $q$ are two roots of the equation $x^2+7x+3=0$. If $\frac{3p}{1-2p}$ and $\frac{3q}{1-2q}$ are the roots of $lx^2+mx+n=0$ and the greatest common divisor of $l, m, n$ is $1$,then $l-m+n=$

For $0 < c < b < a$,let $(a+b-2c)x^2 + (b+c-2a)x + (c+a-2b) = 0$ and $\alpha \neq 1$ be one of its roots. Then,among the two statements:
$(I)$ If $\alpha \in (-1, 0)$,then $b$ cannot be the geometric mean of $a$ and $c$.
$(II)$ If $\alpha \in (0, 1)$,then $b$ may be the geometric mean of $a$ and $c$.

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4+x^2+1=0$,then $\frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=$

If $\alpha, \beta$ are the real roots of $x^2+p x+q=0$ and $\alpha^4, \beta^4$ are the roots of $x^2-r x+s=0$,then the equation $x^2-4 q x+2 q^2-r=0$ has always

If the extreme value of $3x - 2x^2 + 1$ is $k$, then the set of all real values of $x$ for which $kx^2 + 2x + 1 > 0$ is