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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 $u+v$??


Then how to prove that only 3 set of value of u and v suit y=u+v and 3uv+p=0?

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Have you tried with numbers on Cardano's method? Perhaps doing an example would tell you. –  mixedmath Mar 16 '12 at 13:43
One suggestion: your 0% acceptance rate will lower the amount of responses your questions get, and the quality of the responses. –  Álvaro Lozano-Robledo Mar 16 '12 at 15:03

3 Answers 3

up vote 5 down vote accepted

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 $v$.

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

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

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