Prove that $d$ must be a perfect square 
Prove that if $a,b,c,d$ are integers such that $$(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2 = d,$$ then $d$ is a perfect square.

In order for $(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2$ to be an integer, either $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ has to be the square root of a positive integer or $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ must be a perfect square. How do we deal with those conditions?
 A: Theorem: Suppose that $a+2^{1/3}b+2^{2/3}c=0$ where $a,b,c\in\mathbb{Z}$. Then $a=b=c=0$.
Proof: Note that
$$
\begin{align}
&\left(a+2^{1/3}b+2^{2/3}c\right)\left(\left(a^2-2bc\right)+2^{1/3}\left(2c^2-ab\right)+2^{2/3}\left(b^2-ac\right)\right)\\
&=a^3+2b^3+4c^3-6abc\tag{1}
\end{align}
$$
Thus, if $a+2^{1/3}b+2^{2/3}c=0$, then $(1)$ says that
$$
a^3+2b^3+4c^3-6abc=0\tag{2}
$$
If it is not the case that $a=b=c=0$, we can assume that $\gcd(a,b,c)=1$.
Looking at $(2)$ mod $2$, we see that $\phantom{2}a^3\equiv0\pmod2$; therefore $a\equiv0\pmod2$.
Looking at $(2)$ mod $4$, we see that $2b^3\equiv0\pmod4$; therefore $b\equiv0\pmod2$.
Looking at $(2)$ mod $8$, we see that $4c^3\equiv0\pmod8$; therefore $c\equiv0\pmod2$.
Therefore, $2\mid\gcd(a,b,c)$. Contradiction. Thus, $a=b=c=0$.
QED

Answer to the Question
Suppose that
$$
\left(a+2^{1/3}b+2^{2/3}c\right)^2=d\tag{3}
$$
Expanding $(3)$ gives
$$
\left(a^2+4bc-d\right)+2^{1/3}\left(2ab+2c^2\right)+2^{2/3}\left(b^2+2ac\right)=0\tag{4}
$$
The Theorem guarantees that $2abc+2c^3=0$ and $b^3+2abc=0$. That is, $2c^3=b^3$. Since $2^{1/3}\not\in\mathbb{Q}$, we must have that $b=c=0$.
The Theorem also guarantees that $a^2+4bc-d=0$. Since $b=c=0$, this says that $d=a^2$, which is a perfect square.
A: We expand and get
$$
d=(a+2^{1/3}b+2^{2/3}c)^2\\
=(a^2+4bc)+(2c^2+2ab)2^{1/3}+(b^2+2ac)2^{2/3}
$$
If $d$ is to be an integer, then from the expression above we must have $2c^2+2ab=b^2+2ac=0$. Assuming $a\neq0$, this means that
$$
b=-\frac{c^2}{a}\\
c=-\frac{b^2}{2a}
$$
Inserting one of these into the other, we get
$$
b=-\frac{\frac{b^4}{4a^2}}{a}=-\frac{b^4}{4a^3}\\
b(b^3+4a^3)=0
$$
so since $a\neq0$, and $\sqrt[3]4\notin \Bbb Q$, we must have $b=0$, which gives $c=0$, and $d=a^2$, a perfect square.
On the other hand, if $a=0$, we have $c^2=b^2=0$, so $d=0$. Whether you count this as a perfect square is up to your definitions, but I would say that it is.
A: Here's an alternative approach to the above answer if you know some field theory:
Let $\alpha = a + 2^{1/3}b + 2^{2/3}c$. We want to show that $\alpha$ is rational, from which it follows that $\alpha$ is integer since $\alpha^2 = d$ is integer. Assume otherwise. The field $\mathbb{Q}(2^{1/3})$ has degree $3$ over $\mathbb{Q}$, Since $\alpha$ is not rational and $\alpha \in \mathbb{Q}(2^{1/3})$, $\mathbb{Q}(\alpha)$ must have degree $3$ over $\mathbb{Q}$. Since $\alpha^2 = d$ is integer, the polynomial $f(x) = x^2 - \alpha^2$ is in $\mathbb{Q}[x]$. Now $f(\alpha) = 0$, so $\mathbb{Q}(\alpha)$ has degree at most $2$. This is a contradiction, so $\alpha$ must be rational.
