Can the nested radicals in this expression be simplified?

Question

I have the following expression:

$$ \sqrt{ 1 X^2 - \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} - \sqrt[3]{X^2 Y - \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} } + \sqrt{ 2 X^2 + \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} + \sqrt[3]{X^2 Y - \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} + \frac{2 X^3}{\sqrt{ X^2 - \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} - \sqrt[3]{X^2 Y - \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} }} } - 3 X $$

where both $X$ and $Y$ are always greater than or equal to zero.

Is there any way to simplify it so that it uses fewer nested radicals? I'd particularly like to get all the radicals out of the denominators, if possible. I'd also like to use only real numbers and avoid complex numbers, if possible.

Answer 1

One approach to simplifying radicals is to get repeated expressions out of the way to see things more clearly.

I. Simplification

So the expression shortens to,

$$ Z =\sqrt{ X^2 -\color{blue}v } + \sqrt{2 X^2 + \color{blue}v + \frac{2 X^3}{\sqrt{ X^2 - \color{blue}v}} } - 3 X $$

where $v$ is the cubic root,

$$\color{blue}v = \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} + \sqrt[3]{X^2 Y - \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}}$$

In other words, it seems the OP is trying to solve a quartic.

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To re-construct the quartic, the 1st step is to multiply the four conjugates and collect the variable $Z$,

$$P(Z)= \left(Z -\Big(\sqrt{ X^2 -v } + \sqrt{2 X^2 + v + \tfrac{2 X^3}{\sqrt{ X^2 - v}}} - 3 X\Big)\right)\\ \left(Z -\Big(\sqrt{ X^2 -v } - \sqrt{2 X^2 + v + \tfrac{2 X^3}{\sqrt{ X^2 - v}}} - 3 X\Big)\right)\\ \left(Z -\Big({-\sqrt{X^2 -v}} + \sqrt{2 X^2 + v - \tfrac{2 X^3}{\sqrt{ X^2 - v}}} - 3 X\Big)\right)\\ \left(Z -\Big({-\sqrt{X^2 -v}} - \sqrt{2 X^2 + v - \tfrac{2 X^3}{\sqrt{ X^2 - v}}} - 3 X\Big)\right)=0 $$

hence,

$$P(Z) = Z^4 + 12 X Z^3 + 48 X^2 Z^2 + 64 X^3 Z + \frac{4v^3}{v-X^2} = 0$$

or simply,

$$\color{blue}{P(Z) = Z(Z+4X)^3\,(v - X^2) + 4 v^3 =0}\tag1$$

The 2nd step, re-construct its resolvent cubic,

$$P(v) = \prod_{k=0}^2\left(v-\Big(\sqrt[3]{X^2 Y + \sqrt{D}}\,\zeta_3^k+ \sqrt[3]{X^2 Y - \sqrt{D}}\,\zeta_3^{2k}\Big)\right) = 0$$

with discriminant $D=\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3$ and cube root of unity $\zeta_3 = e^{2\pi i/3}$ to get,

$$\color{blue}{P(v) = v^3 + 2 v Y - 2 X^2 Y=0}\tag2$$

The 3rd step, using resultants in Wolfram Alpha to eliminate $v$ between $(1)$ and $(2)$ yields,

$$Z^4 + 12X Z^3 + 48X^2 Z^2 + 64X^3 Z - 8 Y = 0$$

or more elegantly, the final quartic,

$$Z (Z + 4 X)^3 = 8 Y$$

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II. Hypergeometric solution

Because of the special properties of the quartic,

$$Z (Z + 4 X)^3 = 8 Y$$

if a simple closed-form is wanted (and not necessarily in radicals), then it has a rather neat "cubed" generalized hypergeometric solution,

$$ Z = \frac{Y}{(2X)^3}\left({_3F_2}\Big(\tfrac14,\tfrac24,\tfrac34;\,\tfrac23,\tfrac43;\,-\tfrac{8Y}{27X^4}\Big)\right)^3$$

The OP prefers $X,Y \geq 0$ which implies the cubic discriminant $27X^4+8Y \geq 0$. So for example, let $X = 2$ and $Y = 3$, then the hypergeometric above evaluates as,

$$Z = 0.0460743\dots$$

which is the only positive real root of the quartic.

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III. Euler's solution

If radicals are desired but not in the denominator, then the technique was found by Euler. Given the "depressed" form (which any quartic can be transformed to),

$$w^4+pw^2+qw+r = 0$$

then its four roots are among the 8 sign changes of,

$$w = \frac{\pm\sqrt{u_1}\pm\sqrt{u_2}\pm\sqrt{u_3}}2$$

with the $u_i$ as the three roots of the Euler resolvent cubic,

$$ u^3 + 2p u^2 + (p^2 - 4r)u - q^2 =0$$

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IV. Depressing an equation

As requested by the OP, to depress an equation is to get rid of the $x^{n+1}$ term using a linear transformation. For the general quartic,

$$Z^4+aZ^3+bZ^2+cZ+d = 0$$

let $Z=w+t$, and collect the new variable $w,$

$$w^4 + (\color{blue}{a + 4 t}) w^3 + (b + 3 a t + 6 t^2) w^2 + (c + 2 b t + 3 a t^2 + 4 t^3) w +(d + c t + b t^2 + a t^3 + t^4) = 0$$

then solve $\color{blue}{a+4t = 0}$ for the variable $t$. The $w^3$ term vanishes and we get the depressed form,

$$w^4+pw^2+qw+r = 0$$

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V. Example

Given,

$$w^4+3w^2+2w+1=0$$

then the Euler resolvent is,

$$u^3+6u^2+5u-4=0$$

with two conjugate roots of the quartic,

$$w_1 = \frac{+\sqrt{u_1}+\sqrt{u_2}+\sqrt{u_3}}2 \approx 0.3497 + 1.7469i$$ $$w_2 = \frac{-\sqrt{u_1}-\sqrt{u_2}+\sqrt{u_3}}2 \approx 0.3497 - 1.7469i$$

and similarly for the other pair.

Answer 2

This second answer summarizes five different methods of solving the quartic, since it turns out OP's post is linked to that equation. Each method have their uses. For simplicity, we will use the depressed form,

$$x^4+px^2+qx+r = 0$$

for the first three methods and the same resolvent cubic in $u$,

$$ u^3 + 2p u^2 + (p^2 - 4r)u - q^2 =0$$

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Method 1

Given one real root $u$ of the resolvent cubic, then,

$$x_{1,2} = \frac12\left(\sqrt{u}\pm\sqrt{-(2p+u)-\frac{2q}{\sqrt{u}}}\right)$$ $$x_{3,4} = \frac12\left(-\sqrt{u}\pm\sqrt{-(2p+u)+\frac{2q}{\sqrt{u}}}\right)$$

This factors the quartic into two quadratics and is the form used by Mathematica.

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Method 2

Given all three roots $u_i$ of the resolvent cubic, then,

$$x_k = \frac12\Big({\pm\sqrt{u_1}}\pm\sqrt{u_2}\pm\sqrt{u_3}\,\Big)$$

where the four quartic roots $x_k$ are among the 8 sign changes. This avoids radicals in the denominator.

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Method 3

By using resultants to eliminate $x$ between the depressed quartic,

$$x^4+px^2+qx+r = 0$$

and the special quadratic Tschirnhausen transformation,

$$x^2 + (q/u) x + (p/2) - y = 0$$

where $u$ is a real root of the resolvent cubic. This eliminates the $x^3,x$ terms and the quartic transforms into,

$$y^4+Ay^2+B = 0$$

so can be solved as a "quadratic". The roots of the quartic are among the 8 roots of,

$$x^2 + (q/u) x + (p/2) - y_k = 0$$

for $k=0,1,2,3$. This shows a relation between quadratic and quartic equations.

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Method 4

Again, we use resultants. Eliminate $x$ between the general quartic,

$$x^4+ax^3+bx^2+cx+d = 0$$

and the quadratic Tschirnhausen transformation,

$$x^2+ex+f-y = 0$$

The two variables $e,f$ will suffice to eliminate the $x^3,x^2$ terms and are roots of quadratics. One is left with the "Bring quartic",

$$y^4+Ay+B = 0$$

This has a generalized hypergeometric solution given by,

$$ y = -\tfrac{B}{A}\;{_3F_2}\Big(\tfrac14,\tfrac24,\tfrac34;\,\tfrac23,\tfrac43;\,\tfrac{4^4B^3}{3^3A^4}\Big)$$

and is the only method in the list that does not need a cube root.

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Method 5

We use resultants again. Eliminate $x$ between the general quartic and the cubic Tschirnhausen transformation,

$$x^3+ex^2+fx+g- y = 0$$

where $e,f,g$ suffice to eliminate the $x^3, x^2, x$ terms and are roots of sextics with a solvable Galois group. The quartic transforms into the binomial,

$$y^4-C = 0$$

and can be solved by a single $4$th root extraction. The roots of the quartic are among the 12 roots of,

$$x^3+ex^2+fx+g-C^{1/4}\zeta_4^k = 0$$

for $k=0,1,2,3$.