Proving that a number is an integer. Prove that the following number is an integer:
$$\left( \dfrac{76}{\dfrac{1}{\sqrt[\large{3}]{77}-\sqrt[\large{3}]{75}}-\sqrt[\large{3}]{5775}}+\dfrac{1}{\dfrac{76}{\sqrt[\large{3}]{77}+\sqrt[\large{3}]{75}}+\sqrt[\large{3}]{5775}}\right)^{\large{3}}$$
How can I prove it?

we have $$ \left( 76\, \left(  \left( \sqrt [3]{77}-\sqrt [3]{75} \right) ^{-1}-
\sqrt [3]{5775} \right) ^{-1}+ \left( 76\, \left( \sqrt [3]{77}+\sqrt 
[3]{75} \right) ^{-1}+\sqrt [3]{5775} \right) ^{-1} \right) ^{3}
$$
after expanding we obtain
$$438976\, \left(  \left( \sqrt [3]{77}-\sqrt [3]{75} \right) ^{-1}-
\sqrt [3]{5775} \right) ^{-3}+17328\,{\frac {1}{ \left(  \left( \sqrt 
[3]{77}-\sqrt [3]{75} \right) ^{-1}-\sqrt [3]{5775} \right) ^{2}
 \left( 76\, \left( \sqrt [3]{77}+\sqrt [3]{75} \right) ^{-1}+\sqrt [3
]{5775} \right) }}+228\,{\frac {1}{ \left(  \left( \sqrt [3]{77}-
\sqrt [3]{75} \right) ^{-1}-\sqrt [3]{5775} \right)  \left( 76\,
 \left( \sqrt [3]{77}+\sqrt [3]{75} \right) ^{-1}+\sqrt [3]{5775}
 \right) ^{2}}}+ \left( 76\, \left( \sqrt [3]{77}+\sqrt [3]{75}
 \right) ^{-1}+\sqrt [3]{5775} \right) ^{-3}
$$
computing this we obtain
$$616$$
wow
