Let $x$ and $y$ be two $2-$digit numbers such that $y$ is obtained by reversing the digits of $x$. Suppose they also satisfy $x^2-y^2=m^2$ for some positive integer $m$. The value of $x+y+m$ is

One root of the following given equation $2{x^5} - 14{x^4} + 31{x^3} - 64{x^2} + 19x + 130 = 0$ is

The sum of the cubes of all the roots of the equation $x^{4}-3 x^{3}-2 x^{2}+3 x+1=10$ is

If $\alpha $ and $\beta $ are the roots of the quadratic equation, $x^2 + x\, sin\,\theta  -2sin\,\theta  = 0$, $\theta  \in \left( {0,\frac{\pi }{2}} \right)$ then $\frac{{{\alpha ^{12}} + {\beta ^{12}}}}{{\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right){{\left( {\alpha  - \beta } \right)}^{24}}}}$ is equal to

If $\alpha,\beta,\gamma, \delta$ are the roots of $x^4-100x^3+2x^2+4x+10 = 0$ then $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}$ is equal to :-

Let $r$ be a real number and $n \in N$ be such that the polynomial $2 x^2+2 x+1$ divides the polynomial $(x+1)^n-r$. Then, $(n, r)$ can be