when ring with roots is a field I encountered the following problem in abstract algebra textbook.  To determine if the set
$$\{a+b\times 3^{1/3}+c \times 3^{2/3},\quad a,b,c \in \mathbb{Q}\}$$
is a ring, and, if so, a field as well.
I think it's clearly a ring because the product of two such element is also in the set, and each element has additive inverse in the set, for example.
Based on experimenting, I think it is also a field.  For example,
$$\frac{1}{1-3^{1/3}} = -\frac{1}{2}-\frac{1}{2}3^{1/3}-\frac{1}{2}3^{2/3},$$
which shows that $(1-3^{1/3})$ has multiplicative inverse.
However, I do not know how to prove it.  The interesting this is that if I set up the above problem as a linear system, the determinant of the coefficient matrix is zero, but the solution of course is $a = b = c = -1/2$.  So perhaps linear algebra not the correct way to approach linear combinations like these.
Please help me understand this if you can.

Hint: $\,$ let $\,\omega\,$ be a primitive cube root of unity so that $\,\omega^3=1\,$ and $\,\omega^2+\omega+1=0\,$, then:
$$
(a + b t + ct^2)(a + b\omega t + c\omega^2 t^2)(a + b \omega^2 t + c\omega t^2) = c^3\, t^6 + (b^3 - 3 a b c)\, t^3 + a^3
$$
With $\,t = \sqrt[3]{3}\,$:
$$
\frac{1}{a + b \sqrt[3]{3} + c \sqrt[3]{9}} = \frac{1}{9 c^3 + 3(b^3 - 3 a b c) + a^3}(a + b\omega \sqrt[3]{3} + c\omega^2 \sqrt[3]{9})(a + b \omega^2 \sqrt[3]{3} + c\omega \sqrt[3]{9})
$$

[ EDIT ] $\,$ Worked out example for the inverse of $\,1 - \sqrt[3]{3}\,$ $\iff a=1, b=-1, c=0\,$.

*

*$9 c^3 + 3(b^3 - 3 a b c) + a^3 = 3 \cdot (-1) + 1 = -2$


*$(a + b\omega \sqrt[3]{3} + c\omega^2 \sqrt[3]{9})(a + b \omega^2 \sqrt[3]{3} + c\omega \sqrt[3]{9}) = (1-\omega\sqrt[3]{3})(1-\omega^2\sqrt[3]{3})\\ = 1 - (\omega+\omega^2) \sqrt[3]{3} + \sqrt[3]{9} = 1 + \sqrt[3]{3} + \sqrt[3]{9}$
Then: $\displaystyle\quad
\frac{1}{1-\sqrt[3]{3}} = \frac{1}{-2}\left(1 + \sqrt[3]{3} + \sqrt[3]{9}\right)\,$.
