MHB Find positive integers for both a and b

[anemone]

I have a question relating to solving for both a and b in the following question:

Find positive integers a and b such that:

$\displaystyle \left(\sqrt[3]{a}+\sqrt[3]{b}-1 \right)^2=49+20\sqrt[3]{6}$

This one appears to be tough because it doesn't seem right to expand the left hand side and I have tried that but it lead me to nowhere closer to finding both integers values for a and b.

Also, I've tried to work on the right hand side as my effort revolved around rewriting it as the square of sum of two terms but this has not been a fruitful approach as well.

Do you guys have any idea on how I can solve this one, please?

Thanks in advance.

 

[MarkFL]

By the way, anemone has asked me to add that this problem appears in the British Math Olympiad 1993-2005 (round 2, Wednesday 23 February 2000) and can be found at:

http://cage.ugent.be/~hvernaev/Olympiade/UKMO93-05.pdf

Page 16, problem 3.

 

[Fernando Revilla]

Find positive integers a and b such that:

$\displaystyle \left(\sqrt[3]{a}+\sqrt[3]{b}-1 \right)^2=49+20\sqrt[3]{6}$

One way: expressing $\sqrt{49+20\sqrt[3]{6}}=\alpha +\beta \sqrt[3]{6}+\gamma \sqrt[3]{36}$ with $\alpha,\beta,\gamma\in\mathbb{Q}$ we get $\alpha=-1,\beta=\gamma=2$. Now, identifyng, we obtain $a=48,b=288$ (0r reciprocally).

P.S. For the first step, we have used that a basis of $\mathbb{Q}(\sqrt[3]{6})$ over $\mathbb{Q}$ is $B=\{1,\sqrt[3]{6},\sqrt[3]{36}\}$. I ignore if at Olympiad level the students cover some similar property (of course without using the theory of extension fields).