System of Radical Equations 

Immediate thought: Since both results are integers, all roots should evaluate to integers (not trivially true but a good first guess), thus both answers must be raised to the sixth power (the least common multiple). Which gives us the obvious 4+9 on the first, in other words x=2^6 and y=3^6. Which then checks out on the bottom (8+27).

At time of

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Accepting unrestricted answers: from

With integers as condition, it’s a simpler exercise. Otherwise there should be 3 sets of solutions. Can do another round of substitutions u=a+b and v = ab

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The equations are symmetrical in x and y. So you just have to calculate one solution and say in the end that you can swap the two values, because of symmetry.

Proper work

We can assume that a &b are non nagative integers from the beginning because we interested in x and y which they are non nagative integers

I made u⁶=x and t⁶=y

This is a system of two equations with two unknown entities, which means that there is a solution that can always be found without requiring any additional restriction. The requirement that the solutions be integers is an additional restriction that can only be true by chance, which is the case in this system.

U put x= y⁶. To simplify ig a little but and turn it into a polynomial. With a condition x>0. You should get an equation of 6th defree which since it is above 4th degree in general has no explicit formula for the solution.

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Can you do a video on functional differential equations ?