Find $n$ elements of $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$ linearly independent We know that $\mathbb{R}$ is a linear space over $\mathbb{Q}$, denoted by $\mathbb{R_{\mathbb{Q}}}$,where the vector addition is the real numbers and scalar multiplication is the number in $\mathbb{Q}$ times the number in $\mathbb{R}$.
Then $ \forall n\in\mathbb{N^{+}}$,$\{1,\sqrt[n]{2},\sqrt[n]{2^2},\cdots,\sqrt[n]{2^{n-1}}\}$ is a linearly independent set in $\mathbb{R_{\mathbb{Q}}}.$
Further,we can get $\mathbb{R_{\mathbb{Q}}}$ is  an infinite dimensional linear space.
Let $\mathbb{Q}(\sqrt[3]{2}):=\{a+b\sqrt[3]{2}+c\sqrt[3]{2^2} \Big|a,b,c\in \mathbb{Q}\}.$
Similarly，$\mathbb{R}$ is a linear space over $\mathbb{Q}(\sqrt[3]{2})$, denoted by $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$, where the vector addition is the real numbers and scalar multiplication is the number in $\mathbb{Q}(\sqrt[3]{2})$ times the number in $\mathbb{R}$. We can proof that $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$ also is an infinite dimensional linear space. In fact, If we suppose $\dim(\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}})\in \mathbb{N^{+}},$ then  $\dim(\mathbb{R_{\mathbb{Q}}})\in\mathbb{N^{+}}$ holds.  But forall $n\in\mathbb{N^{+}},$ how can I find $n$ elements of  $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$ such that these are  linearly independent in $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$?

The following may feel like cheating: pick your favorite transcendental element of ${\mathbb R}$, e.g., $\pi$. Then $1, \pi, \pi^2, \dots, \pi^{n-1}$ are linearly independent over the field over algebraic reals and in particular over ${\mathbb Q}(\sqrt[3]{2})$. 
If you want to stick to algebraic elements, look at $1, \sqrt[n]{2}, \dots, \sqrt[n]{2^{n-1}}$ (i.e., your own example) for $n$ that are not divisible by $3$. They form a basis of ${\mathbb Q}(\sqrt[n]{2})$ over ${\mathbb Q}$, but also of ${\mathbb Q}(\sqrt[n]{2},\sqrt[3]{2})$ over ${\mathbb Q}(\sqrt[3]{2})$. 
