${\left[\frac{2}{\sqrt{5x^3+1}-\sqrt{5x^3-1}}\right]}^8+\mathrm{ }{\left[\frac{2}{\sqrt{5x^3+1}+\sqrt{5x^3-1}}\right]}^8$

Rationalizing denominators of both the brackets,

$={\left[\sqrt{5x^3+1}+ \sqrt{5x^3-1 } \right]}^8+ {\left[\sqrt{5x^3+1} - \sqrt{5x^3-1 } \right]}^8$

$=2\left[{}^8{C}_0 {\left(\sqrt{5x^3+1}\right)}^8{+}^8{C}_2{\left(\sqrt{5x^3+1}\right)}^6{\left(\sqrt{5x^3-1}\right)}^2{+}^8{C}_4{\left(\sqrt{5x^3+1}\right)}^4{\left(\sqrt{5x^3-1}\right)}^4{+}^8{C}_6{\left(\sqrt{5x^3+1}\right)}^2{\left(\sqrt{5x^3-1}\right)}^6{+}^8{C}_8{\left(\sqrt{5x^3-1}\right)}^8\right]$

$=2\left[{\left(5x^3-1\right)}^4+28{\left(5x^3+1\right)}^3\left(5x^3-1\right)+70{\left(5x^3+1\right)}^2{\left(5x^3-1\right)}^2+28\left(5x^3+1\right){\left(5x^3-1\right)}^3+{\left(5x^3+1\right)}^4\right]$

So, $n=12$ and 

$m=2\left(5^4+140{.5}^3+70{.5}^4+140{.5}^3+5^4\right)=160000={\left(20\right)}^4$