Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring. Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring.
My professor has gone over how to prove a subring, but I am not sure how to prove a ring. 
This is what I have so far, but I am not sure if it is similar to proving a subring, or what I have left to prove.
To prove this set is a ring, we must prove
(i) $S$ is closed under addition, and
(ii) $S$ is closed under multiplication, and
(iii)$S$ is associative on the operations, and
(iv) $S$ is commutative on the operations, and
(v) $S$ is distributive on the operations, and
(vi) $S$ has an additive identity ($0_R$), and
(vii) All operations in $S$ have an additive inverse.
We will begin by proving part (i). That is, we will prove $S$ is closed under addition. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$ and add it to an arbitrary element $d+e\sqrt[3]{2}+f\sqrt[3]{4} \in S$ to get $(a+b\sqrt[3]{2}+c\sqrt[3]{4})+(d+e\sqrt[3]{2} +f\sqrt[3]{4})$. We will then factor out $\sqrt[3]{2}$ and $\sqrt[3]{4}$ using the distributive property to get $(a+d+(b+e)\sqrt[3]{2}+(c+f)\sqrt[3]{4})$. Since $b,e,c,f \in \mathbb{Q}$, $(b+e),(c+f) \in \mathbb{Q}$, we know $S$ is closed under addition which proves part (i) of the proof. 
We will now prove part (ii). That is, we will prove $S$ is closed under multiplication. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$ and multiply it to an arbitrary element $d+e\sqrt[3]{2}+f\sqrt[3]{4} \in S$ to get $(a+b\sqrt[3]{2}+c\sqrt[3]{4})(d+e\sqrt[3]{2} +f\sqrt[3]{4})$. We will then multiply to get $ad+ae\sqrt[3]{2}+af\sqrt[3]{4}+bd\sqrt[3]{2}+be\sqrt[3]{2}\sqrt[3]{2}+bf\sqrt[3]{2}\sqrt[3]{4}+cd\sqrt[3]{4}+ce\sqrt[3]{2}\sqrt[3]{4}+cf\sqrt[3]{4}\sqrt[3]{4}$. We know $\sqrt[3]{4}=\sqrt[3]{2}\sqrt[3]{2}$, so we can make a substitution to get $ad+ae\sqrt[3]{2}+af\sqrt[3]{2}\sqrt[3]{2}+bd\sqrt[3]{2}+be\sqrt[3]{2}\sqrt[3]{2}+bf\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}+cd\sqrt[3]{2}\sqrt[3]{2}+ce\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}+cf\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}$. We can then simplify this to $ad+ae\sqrt[3]{2}+af(2\sqrt[3]{2})+bd\sqrt[3]{2}+be(2\sqrt[3]{2})+bf(3\sqrt[3]{2})+cd(2\sqrt[3]{2})+ce(3\sqrt[3]{2})+cf(4\sqrt[3]{2})$ factor out $\sqrt[3]{2}$ using the distributive property to get $ad+(ae+bd+af(2)+bd+be(2)+bf(3)+cd(2)+ce(3)+cf(4))\sqrt[3]{2}$. Since $a,b,c,d,e,f,2,3,4 \in \mathbb{Q}$, $(ae+bd+af(2)+bd+be(2)+bf(3)+cd(2)+ce(3)+cf(4)) \in \mathbb{Q}$. Thus, we know $S$ is closed under multiplication which proves part (ii) of the proof.
We will now prove part (vi). That is, we will prove $S$ has an additive identity, or in other words has $0_R$. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$. Since $a,b,c \in \mathbb{Q}$ and $0 \in Q$, we will let $a$, $b$, and $c$ be $0$ to get $0+0\sqrt[3]{2}+0\sqrt[3]{4}$. We can simplify this to get $0+0\sqrt[3]{2}+0\sqrt[3]{4}=0+0+0=0_R$. Since the ring contains 0, we know it has an additive identity and proves part (vi) of the proof.
Can anyone help me get on track, please?

Alternative proof: $\mathbb{Q}[x]$ is a ring and let $I$ be the ideal generated by $x^3 -2$, then we have that $R = \mathbb{Q}[x]/ I$ is a ring. 
We have a homomorphism of rings $\phi: \mathbb{Q}[x] \to \mathbb{R}$ given by $\phi(1) = 1$ and $\phi(x) = \sqrt[3]{2}$, since $\phi(I) \subseteq \{0\}$ this defines a homomorphism $\phi: R \to \mathbb{R}$, now $R$ is generated by $1,x$ and the image of $\phi$ contains, $1, \sqrt[3]{2},\sqrt[3]{2} = \phi(x^2)$ (the image of a map is generated by generators of its domain) so in fact $S = \text{im}(\phi)$, as it is the image of a ring under a homomorphism of rings, it is a ring, a posteriori it is an isomorphism.
