When deriving the Cardano formula from $x^{3}+px+q=0$ we let $x$ be a sum and compare coefficients. So $x=u+v$, then we get a system for $u$ and $v$. We get $(1) -q=u^{3}+v^{3}$ and $(2) u^{3}v^{3}=-(\frac{p}{3})^{3}$ . We solve for $u$ and $v$ using the quadratic formula and a cube root, then add those two expressions together to form $x$.
QUESTION: We have a sum with two nested roots, along with two $\pm$ signs. This should mean the expression yields three results by the changing of the $\pm$ signs. We would have $(+;+), (-;-), (+;-)$ for the $\pm$ signs in the nested roots. But only the $(+;-)$ one works and I don't know why. From the derivations I've seen it was only mentioned that if I substitute the formula to the original $x^{3}+px+q=0$, only the $(+;-)$ one would be valid. Most derivations for the formula don't even talk about this. I could't reproduce this or verify it with any explanation.
So why is it this way? Why only this variant? How do you show that? There's just no info about this online.
We can't change signs here. We just solve the system for $u,v$, which gives us a quadratic equations. Solving it gives:
$u=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}=:\sqrt[3]{t}$
$v=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}=:\sqrt[3]{s}$
Or the other way around, we can obviously interchange $u$ and $v$, they are symmetric. In any case, our $x$ is the sum of these two expressions. Not sure what makes you think we can put the same sign in both expressions. Only thing we can do is switch $u$ with $v$, but their sum doesn't change from that.
The $3$ (not necessary distinct) solutions to our cubic equation come from choosing the $3$ distinct third roots of $t$. For any third root of $t$ there is a corresponding third root of $s$.