A student was trying to solve the cubic \[x^3-2x-1=0.\]

Unfortunately, they thought they were dealing with the quadratic \(x^2-2x-1=0\) and used the quadratic formula. They got the two solutions \(x=1\pm\sqrt{2}\), neither of which is a root of the original cubic.

Could this wrong method ever work?

Can you find a cubic equation \(ax^3+bx+c=0\) where (at least) one of the roots of the corresponding quadratic \(ax^2+bx+c=0\) is also a root of the cubic?