<s>[INST] 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?
 [/INST] 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
$$x= \frac1{t-1} =\frac{\sqrt[3]7-1}{2-\sqrt[3]7}=\sqrt[3]{49}+\sqrt[3]7+3
$$
</s>