If  $\alpha$  and  $\beta (\alpha <\beta )$  are the two roots of the equation $\frac{1-8{\left({\log }_{10}x\right)}^2}{{\log }_{10}x-2{\left({\log }_{10}x\right)}^2}=1,$ then the value of    $\frac{1}{10}(\frac{{({\alpha }^2{\beta }^3+1)}^2}{{\alpha }^4}-30000{\alpha }^4)$ is ……….   

It is given that  $\alpha$  and  $\beta$  are the roots of equation 
 $\frac{1-8{\left({\log }_{10}x\right)}^2}{{\log }_{10}x-2{\left[{\log }_{10}x\right]}^2}=1$
Let,  ${\log }_{10}x=y$
So,  $\frac{1-8y^2}{y-2{\left(y\right)}^2}=1$
 $$\begin{gathered} 1-8y^2=y-2y^2 \\ \Rightarrow -8y^2+2y^2-y+1=0 \\ \Rightarrow -6y^2+-y+1=0 \\ \Rightarrow -6y^2+y-1=0 \\ \Rightarrow 6y^2+y-1=0 \\ \Rightarrow 6y^2+3y-2y-1=0 \\ y=\frac{1}{3}andy=-\frac{1}{2} \end{gathered}$$
Therefore,  ${\log }_{10}x=-\frac{1}{2}and{\log }_{10}x=\frac{1}{3}$
 $\Rightarrow x={10}^{-\frac{1}{2}}orx={10}^{1/3}\Rightarrow x=\frac{1}{{10}^{1/2}}orx={10}^{1/3}$
So,  $\alpha =\frac{1}{{10}^{1/2}}and\beta ={10}^{1/3}$
Now,  ${\alpha }^2{\beta }^3+1=\left(\frac{1}{{10}^{1/2}}\right){\left({10}^{1/3}\right)}^3+1$
 $$\begin{gathered} =\frac{1}{10}\times 10+1\Rightarrow 1+1=2 \\ \therefore {({\alpha }^2{\beta }^3+1)}^2=2^2=4 \end{gathered}$$
Again,  ${\alpha }^4={10}^{-2}$
 $\Rightarrow {\alpha }^4=\frac{1}{100}or\frac{1}{{\alpha }^4}=100$
Now,  $\frac{1}{10}[\frac{{({\alpha }^2{\beta }^3+1)}^2}{{\alpha }^4}-30000{\alpha }^4]$
= $\frac{1}{10}[{\left(2\right)}^{2}100-30000\times \frac{1}{100}]$
= $\frac{1}{10}[400-300]=\frac{100}{10}=10$