I see where you're going with this, expanding (x + y√2)^3 gives x^3 + 3√2x^2y + 6xy^2 + 2√2y^3 = 7 + 5√2

Now if I substitute x = 1 and y = 1 on the LHS I do get the RHS, and this gives 1 + √2 as required, but is there a more systematic way to find the values of x and y?

It reminds me of the technique of using undetermined coefficients in partial fractions, but not sure how to do it in this case...

Thanks again.

Welcome to the forum!

Suppose that there are some values x and y for which ∛(7 + 5√2) = x + y√2. What happens if you cube both sides of that equation?

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