It is known that Cardano's formula for solving cubic equations is not good in the case of positive discriminants. In this case it expresses the solution through cubic roots of complex numbers. Computing such square roots again leads to cubic equations.
Is there another formula expressing roots of a general cubic equation with positive discriminant through roots of some special cubic equation with one parameter $B$, e.g. $x^3+x-B=0$? The parameter $B$ is expected to be expressed as a rational function of the coefficients of the initial equation and the square root of its discriminant $\sqrt{D}$.
Best Answer
The usual algebraic function for solving cubic equations involves inverting the cubic power function $P_3(x) = x^3$. This can be defined as the function which can be developed in power series around t=1 as $$P_\frac{1}{3}(t) = 1 + \frac{1}{3}(t-1) - \frac{1}{9}(t-1)^2 + \cdots,$$ and which can be analytically continued, with a branch cut from 0 to $-\infty$. The roots of the polynomial $x^3-t$ are then $x = P_{\frac{1}{3}}(t)$, $\omega x$ and $\omega^2x$, where $\omega$ is a primitive cube root of unity.
However, we can use instead the Chebyshev polynomial (here normalized to -2 to 2) $C_3(x) = x^3-3x$, which can be inverted as an algebraic function developed in power series around $t=2$ as $$C_\frac{1}{3}(t) = 2 + \frac{1}{9}(t-2) - \frac{2}{243}(t-2)^2 + \cdots,$$ and which can be analytically continued, with a branch cut from $-2$ to $-\infty$. The roots of the polynomial $x^3-3x-t$ are then $C_\frac{1}{3}(t)$, $-C_\frac{1}{3}(-t)$, and $C_\frac{1}{3}(-t)-C_\frac{1}{3}(t)$.
If we want to solve $x^3-ax+b=0$, then if $a=0$ we take ordinary cube roots. Otherwise, we apply the above three functions to $-(\frac{3}{a})^\frac{3}{2}b$ and multiply by $\sqrt{\frac{a}{3}}$. That is, one of the roots will be $\sqrt{\frac{a}{3}}C\frac{1}{3}(-(\frac{3}{a})^\frac{3}{2}b)$, another will be $-\sqrt{\frac{a}{3}}C\frac{1}{3}((\frac{3}{a})^\frac{3}{2}b)$, and the third will be minus the sum of these two.
Just as $P_\frac{1}{3}(t)$ can be computed via transcendental functions as $\exp(\log(t)/3)$, $C_\frac{1}{3}(t)$ can be computed by $2\cosh(\mathrm{arccosh}(t/2)/3)$ or $2\cos(\arccos(t/2)/3)$. This no more makes the Chebyshev cube root solution a solution in transcendental functions than logarithms make the ordinary cube root solution a solution in transcendental functions. Moreover, in most cases solving a solvable polynomial in Chebyshev radicals leads to a neater solution than solving it in ordinary radicals.