If the quadratic equation $(a + b)x^2 + 2bx + 1 = 0$ has real and distinct roots,then a point having coordinates $(b, a)$:

If the roots of $\sqrt{\frac{1-y}{y}}+\sqrt{\frac{y}{1-y}}=\frac{5}{2}$ are $\alpha$ and $\beta$ $(\beta > \alpha)$ and the equation $(\alpha+\beta) x^4-25 \alpha \beta x^2+(\gamma+\beta-\alpha)=0$ has real roots,then a possible value of $\gamma$ is

If $x=2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$,then $x^3-6x^2+6x=$

The equation $\left(x^4+1\right)=\frac{1}{a}(x+1)^4$ is a reciprocal equation:

If $\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of the equation $a x^4+b x^3+c x^2+d x+e=0$ and $\alpha_2, \beta_2, \gamma_2, \delta_2$ are the roots of the equation $e x^4+d x^3+c x^2+b x+a=0$ such that $0 < \alpha_1 < \beta_1 < \gamma_1 < \delta_1$,$0 < \alpha_2 < \beta_2 < \gamma_2 < \delta_2$,$\alpha_1-\delta_2=2$,$\beta_1-\gamma_2=2$,$\gamma_1-\beta_2=4$,and $\delta_1-\alpha_2=4$,then $a+b+c+d+e=$

If $x, y, z$ are real and distinct,then $u = x^2 + 4y^2 + 9z^2 - 6yz - 3zx - 2xy$ is always