Show that there exist no $a, b, c \in \mathbb Z^+$ such that $a^3 + 2b^3 = 4c^3$ 
Find all positive integer solutions of $a^3 + 2b^3 = 4c^3$.

Proof: There don't exist any integer solutions for the give equation.
Proof by the Well Ordering Principle.
Let $d$ be the set of all such possible combinations in form $(a, b, c)$.
By the Well Ordering Principle, we that every nonempty set of non-negative integers always has a smallest element. So, there must be a tuple $(a, b, c)$ such that $\max(a, b, c)$ is the least possible value in $d$.
We also know that $a$, $b$ and $c$ must be even. Since $a$ must be even if $4c^3$ has to be even but we know that $4c^3$ is even, so $a$ is the same. Similarly it can be shown that all $a$, $b$ and $c$ are even.
Now, let $a = 2x$, $b = 2y$ and $c = 2z$.
But then, 
$$a^3 + 2b^3 = 4c^3$$
$$\implies (2x)^3 + 2*(2y)^3 = 4*(2z)^3$$
$$\implies 8x^3 + 16y^3 = 32z^3$$
$$\implies x^3 + 2y^3 = 4z^3$$
So, $(x, y, z) \in d$. But, $\max(x, y, z) < \max(a, b, c)$. But $(a, b, c)$ was supposed to be the smallest such member. We have a contradiction here. There is no smallest $(a, b, c)$ which satisfy this criterion and hence no $(a, b, c)$ at all and $d$ is empty. Hence, there exist no integers $a$, $b$ and $c$ which satisfy  $a^3 + 2b^3 = 4c^3$
I wrote this proof. Is it enough to prove the above question? Am I right? Are there any mistakes in the above proof? If there are any, please help me. 
Do you have any other ways of proving it?

There is another proof using the well ordering principle.
We use the following fact:

Third powers are can only be congruent to $0 $,  $1 $ or  $8 \mod 9 $. 

Now, we see that the left hand side is congruent to $0$, $1$, $2$, $3$, $6$, $7$ or $8 \mod 9$, while the right hand side is congruent to $0, 4$ or $5 \mod 9$. So the only possibility is that $4c^3 \equiv 0 \mod 9$ and $a^3+2b^3 \equiv 0 \mod 9$. The first modular congruence gives that $3 \mid c$. The second can only hold true if $3 \mid a$ and $3 \mid b$. 
If one of those isn't true, then we get either a combination of $0 \mod 9$ and $2$ or $7 \mod 9$, a combination of $0 \mod 9$ and $1$ or $8 \mod 9$, a combination of $1$ or $8 \mod 9$ and $2$ or $7 \mod 9$, none of which add to $0 \mod 9$. 
Now suppose that $(a,b,c)$ is the solution with the smallest $a$. 
We know that $3 \mid c$, $3 \mid a$ and $3 \mid b$. 
Hence write $a=3x, b=3y, c=3z$ and then is $(x,y,z)$ also a solution with smaller $a$. 
