In here: http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method
The Cardano's method says that for $ax^3+bx^2+cx+d=0$, $x=y-b/(3a)$, and $y=u+v$. We have found the value of $u^3$ and $v^3$ at last. But both $u^3$ and $v^3$ should have $3$ roots respectively (including complx), so the value of $u^3+v^3$ should have $3*3=9$ roots?!
So why's there only $3$ roots for $u+v$??
Edit:
Then how to prove that only 3 set of value of u and v suit y=u+v and 3uv+p=0?