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 $(p+q)^2$
 A: 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.
A: I did an Excel search and found $(288,48)=(6^2*2^3,6*2^3)$ and the reverse as solutions.  This is confirmed by Alpha
A: 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 $a$, $b$
A: 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.
