Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano 
Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$

Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.
Also, can we prove that there is only one real solution without using the discriminant?
 A: As an alternative to the Cardano’s method, substituting $x=3+2\sqrt{14}\cosh t$ into $x^3-9 x^2-15x-6=0$ to get
$$4\cosh^3t-3\cosh t = \cosh 3t = \frac{15}{4\sqrt{14}}$$
which leads to $t=\frac13\cosh^{-1}\frac{15}{4\sqrt{14}}$ and, then, the solution
$$x=3+2\sqrt{14}\cosh\left(\frac13\cosh^{-1}\frac{15}{4\sqrt{14}}\right)$$
which is numerically equal to $3+2\sqrt[3]{7}+\sqrt[3]{49}$.
A: Substitute $x=\frac1{t-1}$ to translate $x^3-9 x^2-15x-6 =0$ into
$$6t^3-3t^2-3t-1=0$$
and then rewrite it as $7t^3=(t+1)^3$,
which yields $t=\frac1{\sqrt[3]7-1}$ and in turn the solution
$$x= \frac1{t-1} =\frac{\sqrt[3]7-1}{2-\sqrt[3]7}=\sqrt[3]{49}+\sqrt[3]7+3
$$
A: If $y=x-3$ and $y^3-42y=105$, then take $y=t+r\implies t^3+r^3+(3tr-42)y=105$, then: 
$$3tr-42=0$$$$t^3+r^3=105$$
$$3tr-42=0\implies tr=14\implies t^3r^3=2744$$
Then $t^3$, and $r^3$ are roots of:
$$Z^2-105Z+2744=0$$
Then $Z_1=49$ and $Z_2=56$, $\implies t^3=49$ and $r^3=56$, $\implies t=\sqrt[3]{7^2}$ and $r=2\sqrt[3]{7}$
A: When you say "full Cardano" are you referring to using a formula? Because there is a process for solving a cubic that is constructive. But maybe you do not want this because it is essentially how to construct Cardano's formula.
Substituting $x=y+3$ leads to:
$$\begin{align}
x^3-9x^2-15x-6&=0\\
(y+3)^3-9(y+3)^2-15(y+3)-6&=0\\
y^3-42y-105&=0
\end{align}$$
Now introduce parameters $u$ and $v$ such that their sum is a solution $y$:
$$\begin{align}
(u+v)^3-42(u+v)-105&=0\\
u^3+v^3+(3uv-42)(u+v)-105&=0
\end{align}$$
We have freedom to choose $u$ such that $3uv-42=0$. That implies $v=\frac{14}{u}$ and simplifies the equation above:
$$\begin{align}
u^3+\left(\frac{14}{u}\right)^3-105&=0\\
u^6-105u^3+14^3&=0
\end{align}$$
This is quadratic in $u^3$:
$$\begin{align}
u^3&=\frac{105\pm\sqrt{105^2-4\cdot14^3}}{2}\\
u^3&=\frac{105\pm7}{2}\\
\end{align}$$
You may assume the "$+$" for $u^3$, since the "$-$" would give the symmetric $v^3$.
$$\begin{align}
u^3&=\frac{105+7}{2}=56\\
u&=2\sqrt[3]{7}
\end{align}$$
From which $v=\frac{14}{2\sqrt[3]{7}}=\frac{7}{\sqrt[3]{7}}=\sqrt[3]{7^2}$.
And now $x=y+3=u+v+3=2\sqrt[3]{7}+\sqrt[3]{7^2}+3$.
