(From Kirwan's 'Complex Algebraic Curves', chapter 2 q10. Had a search around but can find no solution!)
Show there's a projective transformation taking these points to the points $$\begin{array} \\ [0,1,-1] & [-1,0,1] & [1,-1,0] \\ [0,1,\alpha]& [\alpha,0,1] & [1,\alpha,0] \\ [0,\alpha,1] & [1,0,\alpha] & [\alpha,1,0] \end{array} $$
with ${\alpha^2 - \alpha + 1 = 0}$. Show also that a projective curve of degree 3 passes through these nine points if and only if it is defined by a polynomial of the form
${x^ 3 + y^3 + z^3 + 3\lambda xyz}$
for some ${\lambda \in \mathbb C} $ , and it is singular precisely when
${\lambda \in \{\infty,-1,\alpha,\bar\alpha\} }$
in which case it is the union of 3 lines in $\Bbb{P}^2 .$
$\mathbf{Attempt:}$
My main issue is the first part. I was hoping to show that the nine points could be covered by three lines not all meeting in any one point, which I then showed meant a projective transformation could take those three lines to {x = 0} , {y = 0}, {z = 0}. Then I could get the points to bunch into threes, each set of three having a zero coordinate in one place. The equation in \alpha says it's a cube root of -1, but I'm not sure how come it needs to be. I can see that a curve defined by such a polynomial will hit all of those nine points, and the 'union of three lines' bit is fine. But as for these being \mathbf{all} of the curves passing through all nine points, I'm again stuck.
Would be ever so grateful if someone could explain this to me.