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?

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.
