Proving that a number is an integer.

Hint :

$\sqrt[3]{75}=a$

$\sqrt[3]{77}=b$

$\sqrt[3]{5775}=ab$

$(a^3+b^3)/2=76$

Next , try to simplify expression .

Following the simplifications suggested by MathBot, we have: $$\left( \dfrac{76}{\dfrac{1}{b-a}-ab}+\dfrac{1}{\dfrac{76}{b+a}+ab}\right)^{\large{3}}$$ Let's just take the part inside the parentheses, and put it over a common denominator. $$\dfrac{\dfrac{76^2}{b+a} + 76 ab + \dfrac{1}{b-a} - ab}{\left(\dfrac{1}{b-a}-ab\right)\left(\dfrac{76}{b+a}+ab\right)}$$ Expand the denominator: $$\dfrac{\dfrac{76^2}{b+a} + 75 ab + \dfrac{1}{b-a}}{\dfrac{76}{b^2-a^2}+\dfrac{ab}{b-a}- \dfrac{76ab}{b+a}-a^2b^2}$$ Put the numerator and denominator on a common denominator: $$\dfrac{\dfrac{76^2b - 76^2 a + 75 a b^3 - 75 a^3 b + b + a}{b^2-a^2}}{\dfrac{76 + ab^2 + a^2 b - 76 ab^2 + 76a^2 b - a^2 b^4 + a^4 b^2}{b^2-a^2}}$$ Simplify: $$\dfrac{76^2b - 76^2 a + 75 a b^3 - 75 a^3 b + b + a}{76 - 75 ab^2 + 77a^2 b - a^2 b^4 + a^4 b^2}$$ Remember that $a^3 = 75$ and $b^3 = 77$: $$ \dfrac{(76^2 - 75^2 + 1)b + (75 \cdot 77 - 76^2 + 1) a}{76 - 75 ab^2 + 77a^2 b - 77 a^2 b + 75 a b^2} \ = \ \dfrac{152b + 0 a}{76} \ = \ 2b.$$ Remember we need to cube the whole thing, so the answer is $(2b)^3 = 8 b^3 = 8\cdot 77$, or $616$.

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

The conjugate expression of $\sqrt[3]{a} \pm\sqrt[3]{b}$ is $\sqrt[3]{a^2} \mp\sqrt[3]{ab}+\sqrt[3]{b^2} $. You can use that to rationalise the denominators. The expression inside the parentheses is $\sqrt[3]{77}$ so that finally you get $77$.