I'm reading Cassels' Lectures on Elliptic Curves. I have trouble with Lemma 1, chapter 18:
Let $a,b,c$ distinct integers $>1$ and $d=abc$ cube-free. Let $u,v,w \in \mathbb{Z}$ not all $0$ such that $au^3+bv^3+cw^3=0$. Then there are $x,y,z \in \mathbb{Z}$ such that $x^3+y^3+dz^3=0$.
He defines $\rho$ to be a primitive cubic root of $1$, $\epsilon=au^3+\rho bv^3+\rho^2 cw^3$ and $\eta=au^3+\rho^2 bv^3 +\rho cw^3$ and proves that
$$\epsilon+\eta=3au^3$$ $$\rho \epsilon + \rho^2 \eta=3cw^3$$ $$\rho^2 \epsilon + \rho \eta=3bv^3$$
Multiplying this three values one obtains $\epsilon^3+\eta^3=-d\xi^3$ where $\xi=-3uvw$. So the points $(\epsilon, \rho \eta, \xi)$ and $(\eta, \rho^2 \epsilon, \xi)$ are conjugates over $\mathbb{Q}$. Then he states that the line joining them meets $X^3+Y^3+dZ^3=0$ in a rational point different from $(1,-1,0)$.
I can't understand his last affermation. How can I see that there is another point? Thanks!
As suggested in the comments, it suffices to verify that the three points $$(\epsilon,\rho\eta,\xi),\qquad(\eta,\rho^2\epsilon,\xi),\qquad(1,-1,0),$$ are not collinear. This can be shown by noting that the determinant $$\left|\begin{matrix} \epsilon&\rho\eta&\xi\\ \eta&\rho^2\epsilon&\xi\\ 1&-1&0\end{matrix}\right| =\xi(\rho\eta-\eta+\epsilon-\rho^2\epsilon)=9(\rho-\rho^2)cuvw^4\tag{1},$$ is nonzero; the last equality comes from plugging in the expressions for $\epsilon$, $\eta$ and $\xi$ as follows: \begin{eqnarray*} \rho\eta-\eta+\epsilon-\rho^2\epsilon&=&\rho(au^3+\rho^2bv^3+\rho cw^3)-(au^3+\rho^2bv^3+\rho cw^3)\\ &\ &+(au^3+\rho bv^3+\rho^2cw^3)-\rho^2(au^3+\rho bv^3+\rho^2cw^3)\\ &=&(bv^3+au^3\rho+cw^3\rho^2)-(au^3+cw^3\rho+bv^3\rho^2)\\ &\ &+(au^3+bv^3\rho+cw^3\rho^2)-(bv^3+cw^3\rho+au^3\rho^2)\\ &=&(bv^3-au^3+au^3-bv^3)\\ &\ &+(au^3-cw^3+bv^3-cw^3)\rho\\ &\ &+(cw^3-bv^3+cw^3-au^3)\rho^2\\ &=&(au^3-cw^3+bv^3-cw^3)(\rho-\rho^2)\\ &=&(au^3+bv^3+cw^3-3cw^3)(\rho-\rho^2)\\ &=&-3cw^3(\rho-\rho^2). \end{eqnarray*} You already note that $\xi=-3uvw$ and hence identity $(1)$ above follows. By assumption $c\neq0$ and hence the deteriminant is nonzero if and only if $uvw^4\neq0$, i.e. if and only if $u$, $v$ and $w$ are all nonzero.
Suppose toward a contradiction that $uvw=0$; by symmetry, without loss of generality we may assume that $w=0$ and $a>b$. Then $au^3+bv^3=0$ and without loss of generality $u$ and $v$ are coprime; note that $uv\neq0$ because $u$, $v$ and $w$ are not all $0$ by assumption. Then $$1<\frac ab=-\frac{v^3}{u^3}=\left(-\frac vu\right)^3,$$ which shows that $|v|>1$ and that, contrary to the given assumptions, the integer $$d=abc=\frac abb^2c=\left(-\frac vu\right)^3b^2c=v^3\left(-\frac{b^2c}{u^3}\right),$$ is not cube-free as it is divisible by $v^3$. We conclude that the determinant is indeed nonzero, and hence that the constructed rational point on the curve $X^3+Y^3+dZ^3=0$ is different from the point $(1,-1,0)$.