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. 

We do not need to account for your second case, since it is trivially impossible due to the fact that for positive integers $ a, b, c, d $ the left hand side is positive and the right hand side is negative. Your first case is settled by the following theorem:
Theorem. Let $ a, b \in \mathbf Q^{\times} $ be non-perfect cubes such that $ a/b $ is not a perfect cube in $ \mathbf Q $. Then, $ \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{c} + \sqrt[3]{d} $ for $ c, d \in \mathbf Q $ if and only if $ \{ a, b \} = \{ c, d \} $ (i.e the solution is unique up to permutation).
Proof. Write $ b = qa $ for $ q \in \mathbf Q $ not a perfect cube. Then, we have an equality
$$ \sqrt[3]{a} + \sqrt[3]{qa} = \sqrt[3]{c} + \sqrt[3]{d} $$
These cube roots all lie in some field extension $ L $ of $ \mathbf Q $, and we can talk about the field trace $ T = \textrm{Tr}_{L/\mathbf Q} $. Since $ x^3 - a $ is irreducible in $ \mathbf Q[x] $ if $ a $ is not a perfect cube, and the trace is a multiple of the sum of conjugates, we find that the trace of all irrational cube roots of rational numbers vanish. This is the key idea of the proof. Multiply both sides by $ a^{2/3} $ to get
$$ a + a\sqrt[3]{q} = \sqrt[3]{a^2 c} + \sqrt[3]{a^2 d} $$
If neither cube root on the right hand side was rational, taking traces would yield the absurd $ a = 0 $. Therefore, at least one of $ a^2 c, a^2 d $ is a perfect cube. Assume that it is $ a^2 c = u^3 $ without loss of generality, and now multiply both sides of the original equation by $ (qa)^{2/3} $ to get
$$ a \sqrt[3]{q^2} + qa = \sqrt[3]{q^2 a^2 c} + \sqrt[3]{q^2 a^2 d} $$
By the same argument, one of the cube roots on the right hand side must be rational, since $ q^2 $ is not a perfect cube. If it was $ q^2 a^2 c $, then $ q^2 $ would also be a perfect cube as the ratio of two perfect cubes, which is a contradiction. Therefore $ q^2 a^2 d = v^3 $ is a perfect cube. Substituting into the original equation gives
$$ \sqrt[3]{a} + \sqrt[3]{qa} = \frac{u \sqrt[3]{a}}{a} + \frac{v\sqrt[3]{qa}}{qa} $$
However, $ \sqrt[3]{a} $ and $ \sqrt[3]{qa} $ are linearly independent over $ \mathbf Q $, which means that the only vanishing linear combination is the trivial one, and we have
$$ \frac{u}{a} = \frac{v}{qa} = 1 $$
Now, $ u^3 = a^2 c $ gives $ c = a $, and $ v^3 = q^2 a^2 d $ gives $ d = qa = b $. QED.
Now, since $ 288/48 = 6 $ is not a perfect cube in $ \mathbf Q $ (and neither are $ 288 $ and $ 48 $), the result follows.
