Question Regarding Cardano's Formula In Cardano's derivation of a root of the cubic polynomial $f(X)=X^3+bX+c$ he splits the variable $X$ into two variables $u$ and $v$ together with the relationship that $u+v=X$. From this he finds that $x=u+v$, where
$$u=\sqrt[3]{\frac{-c}{2}+\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}$$ and $$v=\sqrt[3]{\frac{-c}{2}-\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}$$
is a root of $f(X)$. 

Is there an intuitive explanation for why Cardano splits the variable $X$ into two parts, $u$ and $v$?


Cardano knew that any quadratic equation of the form $$x^2+bx+c=0\tag{1}$$ can be written as $$x^2-(u+v)x+uv=0,\tag{2}$$ where $u$ and $v$ are the roots  of the equation. Since by setting $t=u+v$ in the reduced cubic $$t^3+pt+q=0\tag{3}$$ we get $$(u^3+v^3+q)+(3uv+p)(u+v)=0,\tag{4}$$ then every root of the system $$u^3+v^3+q=0\tag{5a}$$  $$3uv+p=0\tag{5b}$$ is a root of $(4)$ as well, and based on the property of the quadratic equation indicated in $(2)$ it's now easy to find a formula for $t$ satisfying equation $(3)$. 
Added. We just need to find two numbers $u^3$ and $v^3$ such that their sum is $-q$ and their product is $-p^3/27$, which we know from $(1)-(2)$ are the roots of the quadratic equation $$Y^2+qY-\frac{p^3}{27}=0.\tag{6}$$
Consequently,
$t=u+v=\sqrt[3]{u^3}+\sqrt[3]{v^3}$.
