Show that $\sqrt[3]{1+\sqrt{3}}$ isn't an element of the field $\mathbb{Q}(\sqrt{3} ,\sqrt[3]{2})$ Setting $\alpha = \sqrt[3]{2}$ and $a+b\alpha+c\alpha^2=\sqrt[3]{1+\sqrt{3}}$ for $a,b,c$ in $\mathbb{Q}(\sqrt{3})$ (The minimal polynomial for $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{3})$ is $x^3-2=0$ since if it were reducible that would imply the existence of $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{3})$ which is easily shown not to be true)  . This comes down to proving that the equations $$ a^3+9ab^2=1 $$ and $$ 3b^3+3a^2b=1$$ have no solutions $a,b$ in $\mathbb{Q}(\sqrt{3})$. This is obviously extremely difficult to prove so there should be a better approach.
This is too long for a comment.
Considering the the irreducible polynomials for $\sqrt[3]{2}$ and $\sqrt[3]{1+\sqrt{3}}$ over $\mathbb{Q}(\sqrt{3})$ we see they are $x^3-(1+\sqrt{3})$ and $x^3-2$ respectively. If the two extensions in the comments are equal that would imply that there is an automorphism of the field $\mathbb{Q}(\sqrt{3})$ which fixes $\mathbb{Q}$ and sends $1+\sqrt{3}$ to $2$ but there is only one nontrivial automrphism of this field that fixes $\mathbb{Q}$ and it sends $\sqrt{3}$ to $-\sqrt{3}$. So the two extensions can't be isomorphic, let alone equal. Is this correct?

Here is an alternative solution, which is essentially the same as Lubin's but with a more elementary presentation.
Lemma. Let $K$ be a subfield of $\mathbb C$ with $\sqrt[3]{2}\not\in K$ (whence it easily follows that $[K(\sqrt[3]{2}):K]=3$). Let $k\in K$. Then $k$
is a cube in $K(\sqrt[3]{2})$ iff one of $k,\frac{k}{2},\frac{k}{4}$ is already
a cube in $K$.
Proof of lemma. The "if" direction is obvious, so let us concentrate on the "only if" part : suppose that $k$ is a cube in $K(\sqrt[3]{2})$, so that
$k=(a+b\sqrt[3]{2}+c\sqrt[3]{4})^3$ for some $a,b,c\in K$. Putting $\theta=a+b\sqrt[3]{2}+c\sqrt[3]{4}$, one can compute that
$$\theta^2=(2ac + b^2)\sqrt[3]{4} + 2(ab + c^2)\sqrt[3]{2} + (a^2 + 4bc)$$
and
$$\theta^3=3(ab^2+a^2c+2bc^2)\sqrt[3]{4} + 3(a^2b+2b^2c+2ac^2)\sqrt[3]{2} + (a^3+2b^3+4c^3+12abc)$$
Since $\theta^3=k\in K$, we must have $z_1=ab^2+a^2c+2bc^2=0$ and $z_2=a^2b+2b^2c+2ac^2=0$. Now $az_1-bz_2=3c(a^3-2b^3)$, so either $a^3-2b^3=0$
(in which case $a=b=0$ because $\sqrt[3]{2}\not\in K$) or $c=0$. It easily follows that at most one of $a,b,c$ is nonzero and the lemma is proved.
Using the lemma above with $k=1+\sqrt{3}$ and $K={\mathbb Q}(k)$, we now have to
see if $k,\frac{k}{2},\frac{k}{4}$ are cubes in $K$ or not.
If $k=(a+b\sqrt{3})^3$ with $a,b\in{\mathbb Q}$, we deduce $a^3+9ab^2=1$ and
$3b^3+3a^2b=1$. We can write $a=\frac{u}{q},b=\frac{v}{q}$ where $u,v,q$ are integers. Then $u^3+9uv^2=q^3$ and $3u^3+3u^2b=q^3$. Reasoning modulo $5$, it can easily be checked that the only solution to this system in ${{\mathbb F}_5}^3$ is $(u,v,q)=(0,0,0)$. Returning to $\mathbb Z$, this means that each of $u,v,q$ must be divisible by $5$. But by infinite descent, this clearly implies that the only solution is $(u,v,q)=(0,0,0)$ in $\mathbb Z$ also, contradiction.
Similarly, in the two other cases we obtain a system  where $(0,0,0)$ is the only solution modulo $5$. This finishes the proof.
