TITLE: Understanding Cardano's method
QUESTION [9 upvotes]: I was recently reading about Cardano's method for solving the cubic of the form $t^3 + pt + q = 0 $.
So, you substitute $t$ with two linear variables $t = u+v$. You get the equation $u^3 + v^3 + (3uv +p)(u+v) +q = 0$. Everything is okay thus far. However, then Allan Clark, whose book I was reading, states:

Since we have substituted two variables, $u$ and $v$, in place of the one variable $t$, we are free to require $3uv + p = 0$.

He writes this as if it is the most obvious thing in the world, but I cannot understand why this is true.
So, how do you justify that step?

REPLY [8 votes]: The stroke of genius is putting $t=u+v$.
We have $(u+v)^3= 3uv(u+v)+(u^3+v^3)$, or $$(u+v)^3-3uv(u+v)-(u^3+v^3)=0$$
Compare this with $$t^3 + pt + q=0$$
If $3uv=-p$ and $u^3+v^3=-q$, then $t=u+v$ solves $t^3 + pt + q=0$.