Find the possible values of $a$, $b$ and $c$? Given $(a,\space b,\space c)\in \mathbb Z^3$ and  that $$\sqrt[3]{\sqrt{a}+\sqrt{b}} + \sqrt[3]{\sqrt{a}-\sqrt{b}} = c$$
Find the possible values of $a $, $b$, and $c$.
 A: We try to find all solutions in integers $a,b,c$ for which the square and cube root are unambiguously defined, that is we require $a\ge0$ and $b\ge0$.
Let $u=\sqrt[3]{\sqrt a+\sqrt b}$, $v=\sqrt[3]{\sqrt a-\sqrt b}$, $w=uv=\sqrt[3]{a-b}$. (Note that $u,v,w$ need not be integers).
Note that $u^3+v^3=2\sqrt a\ge0$ implies $u^3\ge -v^3$ and hence $u\ge-v$ and finally $c\ge0$. The case $c=0$ leads to $u=-v$ and hence $a=0$. Then we find the solutions
$$\tag1(0,b,0)\quad\text{with arbitrary }b\ge0.$$
For the rest of the argument we may assume that $c>0$.
Moreover $u^3+v^3=2\sqrt a$ implies 
$$\tag22\sqrt a = u^3+v^3=(u+v)^3-3(u+v)uv=c^3-3cw,$$
hence by isolating $-3cw$ and cubing
$$-c^{27}+6c^9\sqrt a-12c^3a+8a\sqrt a=-27c^3(a-b),$$
i.e.
$$\tag33(3c^9+4a)\sqrt a \in\mathbb Z.$$
Since $c>0$ and $a\ge0$, we have $3c^9+4a\ne0$ and conclude that $a=d^2$ is a perfect square with $d\in\mathbb Z$.
Next observe that
$$4b=(u^3-v^3)^2=u^6-3u^3v^3+v^6\\=(u+v)^6-6uv(u+v)^4+9(uv)^2(u+v)^2-4(uv)^3
\\=c^6-6c^4w+9c^2w^2-4(a-b).$$
Thus $w$ is root of a quadratic and of a cubic rational polynomial, hence is rational, i.e. $a-b$ is a perfect cube, say $a=b+e^3$ with $e\in \mathbb Z$.
With this, $(2)$ becomes
$$\tag42d=c^3-3ce.$$
Note that $u,v$ are roots of 
$$x^2-cx+e = x^2-(u+v)x+uv= 0,$$
i.e. 
$$\tag5u=\frac{c+\sqrt{c^2-4e}}2\quad v=\frac{c-\sqrt{c^2-4e}}2$$
and we require $c^2\ge4e$.
Now let us go backwards:
Select integers $c>0$ and $e\le\frac{c^2}4$ such that $c\equiv0\pmod 2$ or $e\equiv 1\pmod 2$.
Then $a:=\frac{c^2(c^2-3e)^2}4$ is a nonnegative integer.
Set $b:=a-e^3$. Then $b\ge0$ because either $e\le 0$ and then $b\ge a\ge0$; or $e>0$ and then $c^2-3e\ge e>0$, i.e. $b = \frac{c^2(c^2-3e)^2-4e^3}4\ge \frac{c^2e^2-4e^3}4=\frac{(c^2-4e)e^2}4\ge0$.
With these values, $(a,b,c)$ is a solution.
With nice parametrizations depending on the parity of $c$ we thus find for even $c=2m$ and $e=m^2-n$ ($m>0$ ,$n\ge0$):
$$\tag6\begin{matrix} a&=&m^2(m^2+3n)^2,\\
b&=&m^2(m^2+3n)^2-(m^2-n)^3=n(3m^2+n)^2,\\
c&=&2m.\end{matrix}$$
And for odd $c$ and odd $e$ ($c=2m+1$, $e=m(m+1)-2n-1$ with $m,n\ge0$):
$$\tag7\begin{matrix}a&=&(2m+1)^2\left(\frac{m(m+1)}2+2+3n\right)^2,\\
b&=&(2m+1)^2\left(\frac{m(m+1)}2+2+3n\right)^2-\left(m(m+1)-2n-1\right)^3,\\
c&=&2m+1.\end{matrix}$$
The solutions given by $(1),(6),(7)$ are complete.
