Using Cardano's formula to find roots The equation I am considering is 
$$x^{3}-63x=162$$
Here is how far I got:
$$3st=-63$$
$$s=\frac{-63}{3t}=-21t$$
$$s^{3}t^{3}=-162$$
$$(-21t)^{3}-t^{3}=-162$$
$$-9261t^{3}-t^{3}=-162$$
Where would I go from here?
 A: Cardano's method would be to write 
$$
x = t - \left( \frac{-63}{3t} \right)= t+\frac{21}{t}
$$
which gives
$$
t^3 -162 +\frac{9261}{t^3} = 0
$$
and if you multiply thru by $t^3$ and write $t^3 = p$ this gives 
$$p^2-162p+9261 = 0$$
where the solution is 
$$p=81\pm30 i \sqrt{3}$$
and to recover $t$ one has to take the cube root of $p$. 
Because $p$ is complex (and the original coefficients were real) this is the infamous "irreducible case" where all three roots are real and although you can easily express the solutions in terms of sums of the cube roots of complex quantities, finding those cube roots would be intractable.  (This particular equation, as pointed out, is easily solved without the cubic formula because it has integer roots.)
When the cubic has only one real solution, Cardano's formula is much more useful.  For example, consider 
$$
x^3-63x=1620
$$
there, the same steps give $p = 810+3\sqrt{71871}$ and 
$$t = \sqrt[3]{810+3\sqrt{71871}} \approx 11.7307$$ and
$$
x = t +\frac{21}{t} \approx 13.5209
$$
which indeed is the sole real root.
A: The roots are given by the factorization
$$
x^3-63x-162= (x + 6)(x + 3)(x - 9),
$$
and this is much easier than to use Cardano's formula for it (because it is not so obvious from the formulas that the roots are indeed integral).
