How to solve $4x^3-3z^2=y^6$ in positive integers? 
Solve in positive integers
  $$4x^3-3z^2=y^6$$  

We are given that $\gcd (x,y) = \gcd (y,z) = \gcd  (x,z) = \gcd (x,y,z) = 1$.  
I do not have the slightest idea how to even begin this question. Seeing the L.H.S. , all I can think is about the cosine triple angle formula but I don't think that'll help here ☺  
Any help will be appreciated.  
Thanks.
 A: $x = y = z = 1$ is a solution and it is the only positive integer solution. 
This can be proved using either Elliptic curve or Fermat's Last Theorem.
Method 1 - Elliptic curve
Let $X= \frac{12x}{y^2}$ and $Y = \frac{36z}{y^3}$. Whenever $y \ne 0$, we have
$$y^6 - 4x^3 + 3z^2 = 0\quad\iff\quad Y^2 = X^3 - 432$$
The RHS is the equation of an elliptic curve. Let me call it $\mathcal{E}$. 
It is clear for every integer solution of LHS, there is a corresponding rational point $(X,Y)$ on $\mathcal{E}$.
To extract information about $\mathcal{E}$, we feed following commands
to the online CAS MAGMA:
Q<x> := PolynomialRing(Rationals());  
E00:=EllipticCurve(x^3-432);  
E00;  
MordellWeilShaInformation(E00);  
Generators(E00);  

We find $\mathcal{E} : Y^2 = X^3 - 432$ has following properties:


*

*The rank of $\mathcal{E}$ is $0$ - this means $\mathcal{E}$ has at most finitely many rational points.

*The torsion subgroup of $\mathcal{E}$ is isomorphic to $\mathbb{Z}/3\mathbb{Z}$ - this
means $\mathcal{E}$ has two non-trivial rational points.

*$(12,36)$ is a generator of the torsion subgroup - this implies the two non-trivial rational points of $\mathcal{E}$ are $$(X,Y) = (12,36) \;\text{ or } (12,-36)$$


Translate this back in terms of $x,y,z$, this means the integral solutions of 
$$y^6 - 4x^3 + 3z^2 = 0,\quad y \ne 0$$
all has the form
$$x = y^2\quad\text{ and }\quad z = \pm y^3$$
If one further impose the condition that $x,y,z$ are pairwise co-prime positive integers,
this leaves us one and only one possibilities. Namely, $x = y = z = 1$.
Method 2 - Fermat Last Theorem
Let $x,y,z$ be any pairwise co-prime positive integer solution to the equation
$$y^6 - 4x^3 + 3z^2 = 0$$
Let $A = y^3+z$, $B = y^3-z$ and $C = 2xy$, we have
$$A^3 + B^3 = (y^3 + z)^3 + (y^3 - z)^3 = 2y^3(y^6 + 3z^2) = (2xy)^3 = C^3$$
In order not to contradict with Fermat's last Theorem, we need $B = 0$. This means
$y^3 = z$. Together with the requirement $\gcd(y,z) = 1$, we find $y = z = 1$ and so does $x$.
