Let $r$ be a root of the equation $x^2+2x+6=0$. The value of $(r+2)(r+3)(r+4)(r+5)$ is equal to

The least positive integer $x$ satisfying $2^{2010} \equiv 3x \pmod{5}$ is

Let $\phi(x)=\frac{x}{(x^2+1)(x+1)}$. If $a, b$ and $c$ are the roots of the equation $x^3-3x+\lambda=0, (\lambda \neq 0)$,then $\phi(a) \phi(b) \phi(c) =$

If $x = \sqrt[3]{{\sqrt{2} + 1}} - \sqrt[3]{{\sqrt{2} - 1}}$,then ${x^3} + 3x = $

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

The sum of two roots of the equation $x^4-x^3-16x^2+4x+48=0$ is zero. If $\alpha, \beta, \gamma, \delta$ are the roots of this equation,then $\alpha^4+\beta^4+\gamma^4+\delta^4=$