If $\alpha ,\beta$  are roots of the equation$x^2-15x+1=0$ . Then the value of ${\left(\frac{1}{\alpha }-15\right)}^{-2}+{\left(\frac{1}{\beta }-15\right)}^{-2}$  is

The given equation $x^2-15x+1=0$ 

$\alpha ,\beta$  are the roots given equation

So $\alpha +\beta =15,\alpha \beta =1$

if $x=\alpha$ , Then $x^2-15x+1=0$  will be

${\alpha }^2-15\alpha +1=0\Rightarrow \frac{1}{\alpha }-15=-\alpha$  

Squaring on both sides

$\Rightarrow {\left(\frac{1}{\alpha }-15\right)}^2={\alpha }^2$ …… ( 1 )

if $x=\beta$ , Then $x^2-15x+1=0$  will be

${\beta }^2-15\beta +1=0\Rightarrow \frac{1}{\beta }-15=-\beta$  

Squaring on both sides

$\Rightarrow {\left(\frac{1}{\beta }-15\right)}^2={\beta }^2$ …… ( 2 )

adding eq 1 and 2

Now ${\left(\frac{1}{\alpha }-15\right)}^{-2}+{\left(\frac{1}{\alpha }-15\right)}^{-2}$  =$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$

                                                          $=\frac{{\left(\alpha +\beta \right)}^2-2\alpha \beta }{{\left(\alpha \beta \right)}^2}$

                                                         = $\frac{225-2}{1}=223$     ($\because$ $\alpha +\beta =15,\alpha \beta =1$ )

$\therefore$ ${\left(\frac{1}{\alpha }-15\right)}^{-2}+{\left(\frac{1}{\beta }-15\right)}^{-2}$ = 223