$\alpha$, $\beta$ ,$\gamma$  are roots of equatiuon $x^3 -x -1 = 0$ then equation whose roots are $\frac{1}{{\beta  + \gamma }},\frac{1}{{\gamma  + \alpha }},\frac{1}{{\alpha  + \beta }}$ is

Let $f: R \rightarrow R$ be the function $f(x)=\left(x-a_1\right)\left(x-a_2\right)$ $+\left(x-a_2\right)\left(x-a_3\right)+\left(x-a_3\right)\left(x-a_1\right)$ with $a_1, a_2, a_3 \in R$.Then, $f(x) \geq 0$ if and only if

Let $x, y, z$ be positive reals. Which of the following implies $x=y=z$ ?

$I.$ $x^3+y^3+z^3=3 x y z$

$II.$ $x^3+y^2 z+y z^2=3 x y z$

$III.$ $x^3+y^2 z+z^2 x=3 x y z$

$IV.$ $(x+y+z)^3=27 x y z$

If for a posiive integer $n$ , the quadratic equation, $x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right) + .\;.\;.\; + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right) = 10n$ has two consecutive integral solutions, then $n$ is equal to:

If the equation $\frac{{{x^2} + 5}}{2} = x - 2\cos \left( {ax + b} \right)$ has atleast one solution, then $(b + a)$ can be equal to