rewriting the solutions of symbolic 3rd and 4th order polynomials

Question

I have symbolic third and fourth-order polynomials which read.

$F_{3}= x^3+p x +q,$

$F_{4}= x^4+p_{0} x^2 +q_{0} x + r_{0}.$

Using the Mathematica, I can find the roots of these polynomials using

Solve[x^3 + p*x + q == 0, x]

Solve[x^4+p0 x^2 +q0 x + r0 == 0, x]

The solutions for instance for the 3rd order polynomial are \begin{align} \left\{\left\{\lambda \to \frac{\sqrt[3]{\sqrt{3} \sqrt{4 p^3+27 q^2}-9 q}}{\sqrt[3]{2} 3^{2/3}}-\frac{\sqrt[3]{\frac{2}{3}} p}{\sqrt[3]{\sqrt{3} \sqrt{4 p^3+27 q^2}-9 q}}\right\},\left\{\lambda \to \frac{\left(1+i \sqrt{3}\right) p}{2^{2/3} \sqrt[3]{3} \sqrt[3]{\sqrt{3} \sqrt{4 p^3+27 q^2}-9 q}}-\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{\sqrt{3} \sqrt{4 p^3+27 q^2}-9 q}}{2 \sqrt[3]{2} 3^{2/3}}\right\},\left\{\lambda \to \frac{\left(1-i \sqrt{3}\right) p}{2^{2/3} \sqrt[3]{3} \sqrt[3]{\sqrt{3} \sqrt{4 p^3+27 q^2}-9 q}}-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{\sqrt{3} \sqrt{4 p^3+27 q^2}-9 q}}{2 \sqrt[3]{2} 3^{2/3}}\right\}\right\}. \end{align}

Based on the Cardano's Method, solutions of $F_{3}$ can be also written as \begin{align} x &= u+v ,\\ u^{3} &= -\frac{q}{2} + \sqrt{ \frac{q^2}{4} + \frac{p^3}{27}},\\ v^{3} &= -\frac{q}{2} - \sqrt{ \frac{q^2}{4} + \frac{p^3}{27}} . \end{align}

How can I enforce the final solution of Solve in Mathematica to keep radicals in the nominator?

Answer 1

Not that I know a single command that could do such an operation. Here one needs to work a bit. Let us see the example of the cubic equation:

sl = Solve[x^3 + p*x + q == 0, x]


(*  {{x -> -(((2/3)^(1/3) p)/(-9 q + Sqrt[3] Sqrt[4 p^3 + 27 q^2])^(
     1/3)) + (-9 q + Sqrt[3] Sqrt[4 p^3 + 27 q^2])^(1/3)/(
    2^(1/3) 3^(2/3))}, {x -> ((1 + I Sqrt[3]) p)/(
    2^(2/3) 3^(1/3) (-9 q + Sqrt[3] Sqrt[4 p^3 + 27 q^2])^(
     1/3)) - ((1 - I Sqrt[3]) (-9 q + Sqrt[3] Sqrt[4 p^3 + 27 q^2])^(
     1/3))/(2 2^(1/3) 3^(2/3))}, {x -> ((1 - I Sqrt[3]) p)/(
    2^(2/3) 3^(1/3) (-9 q + Sqrt[3] Sqrt[4 p^3 + 27 q^2])^(
     1/3)) - ((1 + I Sqrt[3]) (-9 q + Sqrt[3] Sqrt[4 p^3 + 27 q^2])^(
     1/3))/(2 2^(1/3) 3^(2/3))}}  *)

Let us take the first of its solutions:

sl[[1, 1, 2]]

(*  -(((2/3)^(1/3) p)/(-9 q + Sqrt[3] Sqrt[4 p^3 + 27 q^2])^(
  1/3)) + (-9 q + Sqrt[3] Sqrt[4 p^3 + 27 q^2])^(1/3)/(
 2^(1/3) 3^(2/3))  *)

To better see it I show below the image of the screen:

Here we need to get rid of the radicals in the denominator of the first term.

Step 1:

slA = sl[[1, 1, 2]] // Simplify

(*  (-2 3^(1/3) p + 2^(1/3) (-9 q + Sqrt[12 p^3 + 81 q^2])^(2/3))/(6^(
 2/3) (-9 q + Sqrt[12 p^3 + 81 q^2])^(1/3))   *)

Now we get the radicals in the common denominator.

Step 2:

slB = Hold[Evaluate[Simplify[Numerator[#]*Denominator[#]^2]]]/
   Denominator[#]^3 &[slA]

(*  Hold[6 6^(1/3) (-9 q + Sqrt[12 p^3 + 81 q^2])^(
  2/3) (-2 3^(1/3) p + 
    2^(1/3) (-9 q + Sqrt[12 p^3 + 81 q^2])^(2/3))]/(36 (-9 q + Sqrt[
   12 p^3 + 81 q^2]))  *)

Now we have only the square root (-9 q + Sqrt[12 p^3 + 81 q^2]) in the denominator. Let us separately multiply the numerator and denominator by (9 q + Sqrt[12 p^3 + 81 q^2]):

Step 3:

Hold[Numerator[slB]*(9 q + Sqrt[12 p^3 + 81 q^2])]/
  Simplify[Denominator[slB]*(9 q + Sqrt[12 p^3 + 81 q^2])] // 
  ReleaseHold // ReleaseHold

(*  ((-9 q + Sqrt[12 p^3 + 81 q^2])^(
 2/3) (9 q + Sqrt[12 p^3 + 81 q^2]) (-2 3^(1/3) p + 
   2^(1/3) (-9 q + Sqrt[12 p^3 + 81 q^2])^(2/3)))/(12 6^(2/3) p^3)  *)

yielding this:

You can, of course, transform the obtained expression according to you liking. For example:

Expand[Simplify[ExpandNumerator[slC]]]

(* (-9 q + Sqrt[12 p^3 + 81 q^2])^(1/3)/(
 2^(1/3) 3^(2/3)) - ((3/2)^(2/3) q (-9 q + Sqrt[12 p^3 + 81 q^2])^(
  2/3))/(2 p^2) - (
 Sqrt[12 p^3 + 81 q^2] (-9 q + Sqrt[12 p^3 + 81 q^2])^(2/3))/(
 6 2^(2/3) 3^(1/3) p^2)  *)

returning this:

If you like this expression better, than the initial one then it's done. Have fun!