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.

`$a\in\{-1,0,1\}$`

by a linear substitution (first shift $x$ to get rid of the quadratic term, then scale $x$ to normalize the linear term). $\endgroup$ – Emil Jeřábek supports Monica Oct 7 '11 at 17:48