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?
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
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 $X^3+Y^3+dZ^3=0$: $$ \begin{aligned} x_3 &= (A-B)^3 -9AB^2\ ,\\ y_3 &= (B-A)^3 -9A^2B\ ,\\ z_3 &= z(A^2+AB+B^2)\ .\\[3mm] &\qquad\text{Then:}\\ -dz_3^3 &= -dz^3\cdot (A^2+AB+B^2)^3 \\ & = -(abc)(-3uvw)^3\cdot (A^2+AB+B^2)^3 \\ &= 27\;ABC\cdot (A^2+AB+B^2)^3\ ,\\[3mm] x_3^3 + y_3^3 &= (x_3+y_3)\Big(\ x_3^2 - x_3y_3 +y_3^2 \ \Big)\ ,\\ (x_3+y_3) &=-9AB^2-9A^2B=-9AB(A+B)=9ABC\ ,\\ &\qquad\text{And indeed, after computations:}\\ x_3^2 - x_3y_3 +y_3^2 &= 3(A^2+AB+B^2)^3\ . \end{aligned} $$