Why does cubic equation have 3 roots but not 9?

Question

In here: http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

Cardano's method says that for $ax^3+bx^2+cx+d=0$, $x=y-\dfrac{b}{3a}$, and $y=u+v$. We have found the values of $u^3$ and $v^3$, but both $u^3$ and $v^3$ should have $3$ roots respectively (including complex ones), so $u^3+v^3$ should have $3\times3=9$ roots?!

So why are there only $3$ roots for $u+v$??

Edit:

How do we prove that there are only 3 sets of values of $u$ and $v$ that satisfy $y=u+v$ and $3uv+p=0$?

Answer 1

In Cardano's method (according to the website you point at), there is an imposed condition on the variables $u$ and $v$ given by the equation $3uv+p=0$. Therefore, the value of $u$ determines the value of $v$.

Answer 2

The values of $u$ and $v$ are not independent. In the reduced cubic $y^3+py+q=0$, we have $y=u+v$ and $3uv+p=0$.

Answer 3

Theorem of Lagrange: A polynomial of degree n over a field has at most n roots in that field.