Understanding the proof that $3X^3 + 4Y^3 + 5Z^3 = 0$ has no non-trivial solutions in $\mathbb{Q}$ I am reading Casell's book 'Lectures on Elliptic Curves' and I am trying to understand a Lemma on page 86 that is used to prove that $3X^3 + 4Y^3 + 5Z^3 = 0$ (Selmer's Curve) has no non-trivial solutions in $\mathbb{Q}$, and thus is a counter-example to the Hasse Principle.
I understand everything in the proof apart from the last line. I don't understand why, if you have two points on the curve $X^3 + Y^3 + dZ^3 = 0$ that are conjugate, then it follows the line joining them intersects the curve at a point in $\mathbb{Q}$ that is not $(1,-1,0)$.
Is there some theorem about conjugate points on curves that is used here? Or is there some way to easily show this in this case?
 A: Let us make the computations explicitly when starting with $(u,v,w)$,
thus writing down explicitly a point $(x_3,y_3,z_3)$ algebraically in therms of $(u,v,w)$.

Let $F=\Bbb Q(\rho)$ be the field obtained by adjoining the primitive third root of one.
Let $(u,v,w)$ with $uvw\ne 0$ be a point on the curve $aU^3+bV^3+cW^3=0$.
It is a good idea to introduce
$$
\begin{aligned}
A &= au^3\ ,\\
B &= bv^3\ ,\\
C &= cw^3\ ,\\
0 &= A+B+C\ .
\end{aligned}
$$
and then as in the lemma the point $(x,y,z)$ over $F$ on the curve $X^3+Y^3+dZ^3=0$ with components
$$
\begin{aligned}
x &= A+\rho  \; B +\rho^2\; C\ ,\\
y &= A+\rho^2\; B +\rho\;   C\ ,\\
z &= -3uvw\ne 0\ ,\\[3mm]
&\qquad\text{ it is on the curve because of }\\[3mm]
x+y &= 2A +(\rho+\rho^2)(B+C)=2A+(-1)(-A)=3A\ ,\\
\rho x+\rho^2 y &= 2C +(\rho+\rho^2)(A+B)=2C+(-1)(-C)=3C\ ,\\
\rho^2 x+\rho y &= 2B +(\rho+\rho^2)(A+C)=2B+(-1)(-B)=3B\ ,\\
x^3+y^3 &=(x+y)(\rho x+\rho^2 y)(\rho^2 x+\rho y)=3^3\; ABC=3^3\; \underbrace{abc}_{=d}\;(uvw)^3 =-dz^3\ .
\end{aligned}
$$
The curve $X^3+Y^3+dZ^3=0$ has the $\Bbb Q$-rational point $(1,-1,0)$, so
it is an elliptic curve. Note the symmetric role of $X,Y$ in the equation.
So if $[x':y':z']\sim(x'/z', y'/z')$ is a rational point for it, then $[y':x':z']\sim(y'/z',x'/z')$ is also a rational point.
(We use two coordinates in the affine $(X,Y)$ plane for points with $z\ne 0$.)
We can also "twist" such an $F$-rational point, by replacing $x'$ by $\rho^j x'$ or/and (in the same time)
$y'$ by $\rho^k y'$.
So let us take two points $P,Q$ from this list, and compute $-(P+Q)$.

Digression. The line through $P=[x:y:z]$ and $\bar P=[\bar x:\bar y:\bar z]=[y:x:z]$ in the affine space $(X,Y)$ has the equation
$$
\begin{aligned}
&\left(Y-\frac yz\right) =m\left(X-\frac xz\right)\ ,\qquad\text{ where }
\\
m
&=\frac{y/z-\bar y/z}{x/z-\bar x/z}
=\frac {y-\bar y}{x-\bar x}
=\frac {y-x}{x-y}
=-1\ .
\end{aligned}
$$
So it is a bad idea to use it to produce a "new " point, since the two equations in the affine $(X,Y)$-plane:
$$
\left\{
\begin{aligned}
X+Y &= \frac 1z(x+y)\ ,\\
X^3 + Y^3 &= -d
\end{aligned}
\right.
$$
give only the two intersection points $P,\bar P$, the third one
lives in the projective space and is the "infinity point" (with $Z=0$).
End of Digression.

So let us use the Cassel's points

*

*$P_1=[x:\rho y:z]=[x/z:\rho y/z:1]\sim(x/z,\rho y/z)=(x_1,y_1)$, and

*$P_2=[y:\rho^2 x:z]=[y/z:\rho^2 x/z:1]\sim(y/z,\rho^2 x/z)=(x_2,y_2)$.

The line connecting them has the equation
$$
\begin{aligned}
&\left(Y-y_1\right) =m\left(X-x_1\right)\ ,\qquad\text{ where }
\\
m
&
:=\frac{y_2-y_1}{x_2-x_1}
=\frac{\rho^2 x/z-\rho y/z}{y/z-x/z}
=\frac{\rho^2 x-\rho y}{y-x}
=\frac 
{A(\rho^2-\rho)+C(\rho-\rho^2)}
{B(\rho^2-\rho)+C(\rho-\rho^2)}
\\
&=\frac{A-C}{B-C}\in\Bbb Q\ ,
\\
n &:= y_1-mx_1 
=\frac 1z(\rho y-mx)
=3C\cdot \frac{A-B}{B-C}\in \Bbb Q\ .
\end{aligned}
$$
The corresponding system is
$$
\left\{
\begin{aligned}
Y &= mX + n\ ,\\
X^3 + Y^3 &= -d
\end{aligned}
\right.
$$
We insert $Y$ from the first equation into the second equation, obtain
$$
X^3 +  m^3X^3 +3m^2nX^2 +\dots =0\ .
$$
Two solutions are known, the $X$-component of the used points, which are $x_1=x/z$ and $x_2y/z$,
so the third solution $x_3$ is obtained using Vieta:
$$
x_3=-\frac xz-\frac yz -\frac{3m^2n}{1+m^3}\ .
$$
The corresponding $y_3$ is obtained by inserting $x_3$ in the equation of the used line,
$y_3=mx_3+n$.
Above, note that $-\frac xz-\frac yz=-\frac 1z(x+y)=\frac 1{3uvw}\cdot 3au^ 3 =\frac {au^2}{vw}\in\Bbb Q$.
So $x_3,y_3\in\Bbb Q$. This answers the question.
$\square$

But it may be important to really have a point $(x_3,y_3)$ explicitly.
It turns out, that after the denominators were made clean, the following point is a point on
$X^3+Y^3+dZ^3=0$:
$$
\begin{aligned}
x_3 &= (A-B)^3 -9AB^2\ ,\\
y_3 &= (B-A)^3 -9A^2B\ ,\\
z_3 &= z(A^2+AB+B^2)\ .\\[3mm]
&\qquad\text{Then:}\\
-dz_3^3 &= -dz^3\cdot (A^2+AB+B^2)^3
\\ &
= -(abc)(-3uvw)^3\cdot (A^2+AB+B^2)^3
\\
&= 27\;ABC\cdot (A^2+AB+B^2)^3\ ,\\[3mm]
x_3^3 + y_3^3 &= (x_3+y_3)\Big(\ x_3^2 - x_3y_3 +y_3^2 \ \Big)\ ,\\
(x_3+y_3) &=-9AB^2-9A^2B=-9AB(A+B)=9ABC\ ,\\
&\qquad\text{And indeed, after computations:}\\
x_3^2 - x_3y_3 +y_3^2 &= 3(A^2+AB+B^2)^3\ .
\end{aligned}
$$
