<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] If you don't mind doing a lot of arithmetic, here's an "easy" way to solve things:
Note that $(49+20\sqrt[3]{6})^{3/2} = 788.401\ldots$.  Since you're looking for positive integer solutions $a$ and $b$, one need at worst examine numbers from $1$ to $788$: That is, if $b\ge1$, then $\sqrt[3]{a}\le\sqrt[3]{a}+\sqrt[3]{b}-1 =(49+20\sqrt[3]{6})^{1/2}$, hence $1\le a\le788$.  In doing so, you will presumably stumble across the solution $(48,288)$ and its reverse, and no others.
You can streamline this a bit by looking for solutions with $a\le b$ so that 
$$2\sqrt[3]{a}\le\sqrt[3]{a}+\sqrt[3]{b} = 1+(49+20\sqrt[3]{6})^{1/2}=10.238\ldots$$ 
leads to $a\le134$.  Whether you can easily streamline it further is less clear.
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