How to determine that the solution to the following problem is unique? The question is:
[BMO2 2000 Q3] Find positive integers a and b such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20\sqrt[3]{6}$$   
It suffices to find one solution to gain full marks, which I did by expanding and assuming that neither $a$ nor $b$ are perfect cubes, so that $2\sqrt[3]{ab} = 48$. The solution I get from this method is $(a, b) = (48, 288)$. However, from this two questions arise in my mind:
a) Is this is the only solution? I.e. are there $c$ and $d$ such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = (\sqrt[3]{c} + \sqrt[3]{d} - 1)^2$$
$\Rightarrow  \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{c} + \sqrt[3]{d}$ or $\sqrt[3]{a} + \sqrt[3]{b} - 1 = -\sqrt[3]{c} - \sqrt[3]{d} + 1$
b) Can one solve the original problem in such a way that it is clear that the solutions found are unique?
EDIT: Thank you to gammatester for his remark that there is another solution: $(a, b) = (288, 48)$, but I am referring to distinct solutions which are not permutations of other solutions. 

I apologize, this solution is a bit hand-wavy and (it may in fact be wrong) as my skills/knowledge in this topic is rather weak. Perhaps someone else can later provide a more rigorous version of this argument if it is correct.
We have,
$$\sqrt[3]{a^2}+\sqrt[3]{b^2}-2(\sqrt[3]{a}+\sqrt[3]{b}) = 20\sqrt[3]{6}.$$
Consider the highest power of $6$ that divides $a$ and $b$ modulo $3$. I believe (likely using some argument about the linear independence of cuberoots of integers over $\mathbb{Q}$) that it can be shown that the highest power of $6$ (modulo $3$) that divides $a$ is $6^1$ and similarly $6^2$ for $b$. Such an argument would probably use the fact that the righthand side doesn't contain any $\sqrt[3]{6^2}$ terms.
Also, we have, like you said, 
$$2\sqrt[3]{ab} = 48 = 2^4 \cdot 3$$
$$\implies 2^3 | \sqrt[3]{ab}$$
$$\implies 2^9 | ab.$$
That means that there are $9$ factors of $2$ between $a$ and $b$. Since $a$ contains $1$ factor of $6$ and $b$ contains $2$ factors of $6$, that means that there are a total of $6$ factors of $2$ left between $a$ and $b$ after we subtract out the factors of $2$ belonging to the $6$'s. In order to write $\sqrt[3]{a} = k\sqrt[3]{6}$ and $\sqrt[3]{b} = l\sqrt[3]{6^2}$ for some integers $k,l$; then those $6$ factors of $2$ must be distributed between $a$ and $b$ in multiples of $3$. There are three ways to do this:


*

*Give all six factors of $2$ to $a$

*Give all six factors of $2$ to $b$

*Give both $a$ and $b$ three factors of $2$ each.


This gives us three candidates for our choices of $a,b$:


*

*$a = 6\cdot 2^6$ and $b=6^2$

*$a = 6$ and $b=6^2\cdot 2^6$

*$a = 6\cdot 2^3$ and $b=6^2\cdot 2^3$


Of these three, only the last one solves the equation, hence the solution is unique.
As a side note, I've verified that $(48,288)$ is the only solution up to reordering in Mathematica:
Solve[(x^(1/3) + y^(1/3) - 1)^2 == 49 + 20*6^(1/3) && x > 0 &&   y > 0, {x, y}, Integers]
Hope this helps!
