Expanding the LHS, we get \[ d=a^2+b^2\sqrt[3]{4}+2c^2\sqrt[3]{2}+2ab\sqrt[3]{2}+4bc+2ac\sqrt[3]{4}\hspace{1.5cm}(*)\] Because $ \sqrt[3]{2},\sqrt[3]{4}$ are irrational, the coefficients of each must be 0; that is, \begin{eqnarray} b^2+2ac=0 \\ 2c^2+2ab=0 \end{eqnarray} We wish to show that $ b=c=0$, from which $ d=a^2$, as desired, would follow. Note that $ a\neq 0$ because the RHS of (*) would be irrational. Assume for contradiction that neither $ b$ nor $ c$ are zero. Then, moving the squares of (1) and (2) to one side and dividing, we have \[ \frac{b^2}{c^2}=\frac{2c}{b} \implies 2c^3=b^3 \implies c=b\sqrt[3]{2}\], which contradicts $ b,c\in\mathbb{Z}$. Therefore, at least one of them is zero. Then, (1) implies that the other is also zero, as desired.