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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:

$u = sqrt[3]-fracq2+sqrt(fracq2)^2 + (fracp3)^3$

and

$v = sqrt[3]-fracq2-sqrt(fracq2)^2 + (fracp3)^3$

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:
$x^3 + frac78x- frac2516 = 0$.

$u$ and $v$ are also constrained by $3uv + p = 0$.

In this particular problem you get

$$u = sqrt[3]frac2532 + sqrtfrac1756127648,quad v = sqrt[3]frac2532 - sqrtfrac1756127648$$
so if $alpha$ and $beta$ are real cube roots, all roots are given by

$$u_0 = alpha, quad u_1 = alpha e^frac2pi i3,quad u_2 = alpha e^frac4pi i3$$
$$v_0 = beta, quad v_1 = beta e^frac2pi i3,quad v_2 = beta e^frac4pi i3$$

So $$mathbbR ni -fracp3 = u_kv_j = alphabeta e^frac2pi i3 (k+j)$$

Hence $k+j in 0, 3$, so the valid combinations are $$k = l = 0$$ $$k = 1, quad l = 2$$
$$k = 2, quad l = 1$$

Suppose you have a cubic equation $ax^3+bx^2+cx+d=0$ with a root $x_1$ you found via the Cardano method, or perhaps as a rational root or even a specified root value you used to construct the equation. Then you always have:

$ax^3+bx^2+cx+d=a(x-x_1)(x^2+rx+s)text Eq. 1$

$a(x-x_1)(x^2+rx+s)=ax^3+a(r-x_1)x^2+a(s-rx_1)x-ax_1stext Eq. 2$

So matching the quadratic and constant terms between the right side of Eq. 2 and the left side of Eq. 1 gives

$r=(b/a)+x_1, s=-d/(ax_1)$

With these values of $r$ and $s$ derived, just use the quadratic formula to get the remaining roots:

$x_2,x_3=frac-rpmsqrtr^2-4s2=frac2s-rmpsqrtr^2-4s$