When is $\sqrt[3]{a+\sqrt b}+\sqrt[3]{a-\sqrt b}$ an integer? I saw a Youtube video in which it was shown that
$$(7+50^{1/2})^{1/3}+(7-50^{1/2})^{1/3}=2$$
Since there are multiple values we can choose for the $3$rd root of a number, it would also make more sense to declare the value of this expression to be one of $2, 1 + \sqrt{-6},$ or $1 - \sqrt{-6}$

We may examine this more generally. If we declare $x$ such that 
$$x=(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}$$
$$\text{(supposing } a \text{ and } b \text{ to be integers here)}$$
one can show that
$$x^3+3(b-a^2)^{1/3}x-2a=0$$
Which indeed has $3$ roots.

We now ask 


  For what integer values of $a$ and $b$ is this polynomial solved by an integer?


I attempted this by assuming that $n$ is a root of the polynomial. We then have
$$x^3+3(b-a^2)^{1/3}x-2a$$
$$||$$
$$(x-n)(x^2+cx+d)$$
$$||$$
$$x^3+(c-n)x^2+(d-nc)x-nd$$
Since $(c-n)x^2=0$ we conclude that $c=n$ and we have 
$$x^3+3(b-a^2)^{1/3}x-2a=x^3+(d-c^2)x-cd$$
And - to continue our chain of conclusions - we conclude that
$$3(b-a^2)^{1/3}=d-c^2 \quad\text{and}\quad 2a=cd$$
At this point I tried creating a single equation and got
$$108b=4d^3+15c^2d^2+12c^4d-c^6$$
This is as far as I went.
Use $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ identity. Things will cancel out.
One choice is where $a-b^{1/2}=0$ so $a=4k^3$ for integer $k$ and $b=16k^6$, then
$$(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}=2k$$