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 $\sqrt{D}$.

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    $\begingroup$ The other classic solution for cubics uses trig functions. Then, at least, you can solve a cubic with real coefficients and three real roots without going to the complex numbers. planetmath.org/encyclopedia/ATrigonometricCubicFormula.html $\endgroup$ – Gerald Edgar Oct 7 '11 at 14:26
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    $\begingroup$ ... and $\cos\frac{\arccos B}3$ satisfies $4x^3-3x-B=0$. $\endgroup$ – user2035 Oct 7 '11 at 14:45
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    $\begingroup$ It is a simple exercise to reduce any cubic equation to $x^3+ax+b=0$ with $a\in\{-1,0,1\}$ 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
  • $\begingroup$ To clarify: if we consider the smallest subfield of the real numbers containing the rationals and stable under taking positve $n$th roots of positive numbers, does it contain all real roots of degree 3 rational polynomials? (I expect a negative answer) $\endgroup$ – YCor Dec 29 '15 at 15:56

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.

  • $\begingroup$ Very nice, is there a canonical reference for this? $\endgroup$ – Igor Rivin Sep 9 '12 at 16:59
  • $\begingroup$ I wrote it up for Wikipedia, but that seems to have mutated pretty far; this on the other hand is nice, and I didn't know until just now it had found a home: statemaster.com/encyclopedia/Cubic-equation#Chebyshev_radicals $\endgroup$ – Gene Ward Smith Sep 9 '12 at 17:13
  • $\begingroup$ Great! But does this mean that you thought of it, or is it a classical observation (I see no references, so I assume the first alternative is correct, but do confirm...) $\endgroup$ – Igor Rivin Sep 9 '12 at 18:19
  • $\begingroup$ The specific way I wrote it up is mine, but I think the basic approach so far as just cubics goes and mentioned by Emil Jeřábek traces back to Vieta. The Wikipedia article "Cubic function" cites the Abramowitz and Stegun chapter for "Chebychev cube roots", but I can't find it in there. I've found occasional mentions in odd places of that and "Chebyshev radicals", which likely derive from deleted material on Wikipedia I wrote. Anyway, I think it counts as a folk theorem. $\endgroup$ – Gene Ward Smith Sep 11 '12 at 3:43
  • $\begingroup$ Sorry, looking again I see Jeřábek gave a different normalization, the OPs version is closer. The article on StateMaster derives from stuff I wrote up on Wikipedia and which other people worked on a little also, but which was deleted due to lack of references (the dreaded original research ban.) $\endgroup$ – Gene Ward Smith Sep 11 '12 at 3:53

Another alternative method to solve cubic equation is by root shift method, as explain here: http://eigenmaths.blogspot.my/2015/12/cubic-equation-solved-using-root.html

In the blog, the writer also used the same method to solve quartic equation.

As for the cubic solution using this method, it can be reduced to cardano's form , also explained by the writer here: http://eigenmaths.blogspot.my/2015/12/cubic-equation-continued-transformation.html


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