Prove that $\sqrt{12 x^3-3}$ is irrational First, I thought that it should be easy to prove that $\sqrt{12  x^3-3}$ for $x\neq1$ and $x\in \mathbb{Q}^+$ is irrational, but I am stuck!   
I am almost sure that this is irrational, because if it is not, then some theorems are wrong and it is very improbable.  
Any help or ideas?
 A: Hint $\,\ y^2+3 = 12x^3\,\Rightarrow\, \overbrace{(3+y)^{\large\color{#c00}3}+(3-y)^{\large\color{#c00}3}}^{\Large 18(y^2+3)} = (6x)^{\large\color{#c00}3}\Rightarrow\, y=\pm3, x=1\,$ by $\rm{FLT}_{\large\color{#c00}3}$

Remark $ $ More generally we can completely classify the the points of the Fermat cubic over real quadratic fields using the following result.
Theorem $ $ Solutions of $\,x^3+y^3= 1\,$ over $\,\Bbb Q(\sqrt d)\,$ are in correspondence with the $\,\Bbb Q\,$-points on the elliptic curve  $\,y^2 = x^3 - 432\, d^3$
An elementary proof: Lemma $2.1$ in F. Jarvis and Paul Meekin, The Fermat equation over $\,\Bbb Q(\sqrt 2).$
A: HINT.- You are stuck because your problem is intrinsically difficult. This involves to find the rational points of the elliptic curve $$y^2=12x^3-3$$ You have the obvious point $(x,y)=(1,3)$ which if not of torsion will give infinitely many other rational points in which case the rank of the curve is at least equal to $1$. To see if this is verified we do not need the formulas of adding the group defined by the curve but simply using the old Bachet's method of chords and tangents from which  the group law of elliptic curves comes.
The derivative $y'=\dfrac{18x^2}{\sqrt{12x^3-3}}$ at the point $(1,3)$ is $y'(1)=6$ and the tangent at $(1,3)$ is $y=3(2x-1)$. This tangent cuts the curve at another point given by the system
$$\begin{cases}y=3(2x-1)\\y^2=12x^3-3\end{cases}$$ If this "another" point is distinct of $(1,3)$ we will properly speaking get another point with possibilities of infinitely (not necessarily)  many other points.
The resultant of the system is $$x^3-3x^2+3x-1=(x-1)^3=0$$ 
Thus $(x,y)=(1,3)$ is a torsion point of the elliptic curve $y^2=12x^3-3$ and the rank of it could be  $0$ if does not exist another rational point not of torsion. It seems that in fact the rank is $0$ but, I repeat, this is a difficult, non-elementary problem.
