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?
 A: 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 $616$.
A: 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
A: Hint :
$\sqrt[3]{75}=a$
$\sqrt[3]{77}=b$
$\sqrt[3]{5775}=ab$
$(a^3+b^3)/2=76$
Next , try to simplify expression .
A: 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 $77$.
