This is the second part of a question that asks the same thing but for a quadratic, that part seemed to be fine. The next part asks you to show that: $$x^3-\frac{3}{2}x^2-\frac{3}{2}x+1=0 $$ is the only cubic equation of the form $x^3+px^2+qx+r=0$, where $p,q,r \in \mathbb{R}$ which has the following properties: If $k$ is a (possibly complex) root, then $k^{-1}$ is a root and, if $k$ is a root then $1-k$ is a root.
Now the way I intiallly went about it was to write the general cubic as: $$ x^3+px^2+qx+r=(x-k)(x-\frac{1}{k})(x-(1-k)) $$ And then try to deduce the required equation, though I can't seem to be able to get to it.
A point that confuses me a little is that if we let $k$ be a root then we know that $k^{-1}$ and $1-k$ are also roots. But then couldn't you say that $1-k^{-1}$ must also be a root and consequently so must $1/(1-k^{-1})$. Therefore implying that in order to satisfy the root properties completely there must be 6 roots but obviously this must be wrong.