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 $(p+q)^2$

This one is interesting.
Note that not both $a$ and $b$ are perfect cubes. Why? because otherwise the lhs would be a perfect cube!
Now, let's tackle the case where one of them is a perfect cube, say $a$. In that case, $(\sqrt[3]{a} - 1)$ is an integer. By elementary property of surds, square of this integer must match $49$, integer part of rhs, giving us
$$(\sqrt[3]{a} - 1)^2 = 49$$
And thus, $a = 8^3 = 512$.
The remaining part then reduces to $(7 + \sqrt[3]{b})^2 = 49+20\sqrt[3]{6}$, i.e. $b^{2/3} + 14b^{1/3} = 20(6)^{1/3}$
Edit: This quadratic has no integer solution. I was wrong in assuming $6$ as a solution.
This means neither $a$ nor $b$ is a perfect cube
Edit 2: I first thought integer solutions must somewhere involve integer cubes. Although neither $a$ nor $b$ are perfect cubes, it turns out that $ab$ is a perfect cube and that is a key to the solution. See Calvin's solution to prove that $ab$ is a perfect cube and find $a$, $b$
