<s>[INST] Find $a,b\in\mathbb{Z}^{+}$ such that $\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$ find positive intergers $a,b$ such that
$\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$
Here i tried plugging
$x^3=a,y^3=b$
$(x+y-1)^2=x^2+y^2+1+2(xy-x-y)=49+20\sqrt[3]{6} $
the right hand part is a square hence can be written as $(p+q)^2$
 [/INST] Firstly, if either $a$ or $b$ is a perfect cube, then we have $$ (\sqrt[3]{a}-N)^2 = 49 + 20 \sqrt[3]{6} $$
Convince yourself that the LHS must involve 2 different surd terms, hence this is not possible. Thus, neither $a$ nor $b$ are a perfect cube.
Like you did, expand the terms and consider what they are. Clearly $a, b, a^2, b^2$ are not perfect cubes. If $ab$ is not a perfect cube, then we must have $1 = 49$, which is a contradiction. Hence, we have $ 2\sqrt[3]{ab} + 1 = 49$, or that $ab = 24^3$. We use the substitution $b = \frac{24^3}{a}$ and $x = \sqrt[3]{a}$, and the equation becomes
$$x^2-2x - 48/x + 24^2/x^2 = 20 \sqrt[3]{6}.$$
Multiplying throughout by $x^2$, this equation has real roots of $x = \sqrt[3]{48}$ and $x= \sqrt[3]{288}$, hence $ (a,b) = (48 288) $ and $(288,48)$ are solutions.
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