Given 2 points $ A $ and $ B $ in the complex plane with $ AB=1 $, is it possible to construct with unruled straightedge and compass a point $ C $ on the line $ AB $ such that $ AC \cdot BC^{2}=1 $?
I think this is false. Notice that we must consider the $ 3 $ cases given by the position of $ C $ in respect to $ A $ and $ B $. Suppose that such a point $ C $ were constructible and assume that $ C $ lies between $ A $ and $ B $. Denote $ a=BC $. Then we should have that $ a^{2}(1-a)=1 $, thus $ a^{3}-a^{2}+1=0 $. So the minimal polynomial of $ a $ divides $ X^{3}-X^{2}+1 $ But the polynomial $ X^{3}-X^{2}+1 $ is irreducible over $ \mathbb{Q} $, hence they must coincide. On the other hand, we know that the degree of the minimal polynomial of $ a $ over $ \mathbb{Q} $ must be a power of $ 2 $ as $ a $ is constructible by assumption. Therefore we get a contradiction.
The other two cases, when $ B $ lies between $ A $ and $ C $ as well as when $ A $ lies between $ C $ and $ B $ are treated analogously, by observing that the minimal polynomial of a supposedly constructible number has degree $ 3 $, which is not a power of $ 2 $.
I see no fault in this. Please let me know in case I am wrong. Thank you!