To solve $x^3 + px + q = 0$. I'm using Cardano's method, with the substitution $x = u + v$.
In the special case that $3uv + p = 0$, I've derived $u$ and $v$ as the following:
$u = \sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2 + (\frac{p}{3})^3}}$
and
$v = \sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2 + (\frac{p}{3})^3}}$
The next step is to find $x$, which is $x=u+v$, but how would I go about finding all three values for x when there are 9 possible combinations for $u_1, u_2, u_3,$ and $v_1, v_2, v_3$?
EDIT: If it's of any relevance, this is the expression I'm trying to solve:
$x^3 + \frac{7}{8}x- \frac{25}{16} = 0$.
Best Answer
$u$ and $v$ are also constrained by $3uv + p = 0$.
In this particular problem you get
$$u = \sqrt[3]{\frac{25}{32} + \sqrt{\frac{17561}{27648}}},\quad v = \sqrt[3]{\frac{25}{32} - \sqrt{\frac{17561}{27648}}}$$ so if $\alpha$ and $\beta$ are real cube roots, all roots are given by
$$u_0 = \alpha, \quad u_1 = \alpha e^{\frac{2\pi i}3},\quad u_2 = \alpha e^{\frac{4\pi i}3}$$ $$v_0 = \beta, \quad v_1 = \beta e^{\frac{2\pi i}3},\quad v_2 = \beta e^{\frac{4\pi i}3}$$
So $$\mathbb{R} \ni -\frac{p}3 = u_kv_j = \alpha\beta e^{\frac{2\pi i}3 (k+j)}$$
Hence $k+j \in\{ 0, 3\}$, so the valid combinations are $$k = l = 0$$ $$k = 1, \quad l = 2$$ $$k = 2, \quad l = 1$$