<s>[INST] 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?
 [/INST] 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.
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