Given the root of a cubic equation $Z = \sqrt[3]{Y + \sqrt{Y^2 – \frac{X^6}{27}}} + \sqrt[3]{Y – \sqrt{Y^2 – \frac{X^6}{27}}} – X$ and the assumption that both $X$ and $Y$ are greater than zero, is there a "natural" or "correct" way to find a cubic equation $f(Z) = a Z^3 + b Z^2 + c Z + d$ which has that specific value as its ONLY positive real root? I suspect that the values of $a$, $b$, $c$, and $d$ should all be combinations of only rational numbers and the variables $X$ and $Y$ (no radicals and no irrational, transcendental, or imaginary numbers).
In a previous question of mine, I learned that another expression I have, $Z = \sqrt{X^2 – v} + \sqrt{2 X^2 + v + \frac{2 X^3}{\sqrt{X^2 – v}}} – 3 X$ where $v = \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} + \sqrt[3]{X^2 Y – \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}}$, can be used to find the quartic equation $Z^4 + 12X Z^3 + 48X^2 Z^2 + 64X^3 Z – 8 Y = 0$, or more simply $Z (Z + 4 X)^3 – 8 Y = 0$. I want to do the same thing for the cubic version, if possible.
However, the procedure for finding the quartic involves the fact that there are two separate square radical terms (one of which appears both outside and nested inside the other), and using all four combinations of the positive/negative conjugates of both square radicals to get a total of four roots for the quartic equation in such a way that the radicals all cancel out nicely; and it doesn't seem like the same procedure will work for the cubic case, as there is only a single square radical term that appears twice and replacing it with its conjugate doesn't change the expression.
Am I overlooking something that makes the quartic procedure work for the cubic equation? Is there an alternative way to get at the cubic equation?
More Background
The quartic equation example is connected to a set of values I have for $Z$:
- $Z_0 = Y – 0 X$
- $Z_1 = \sqrt{Y + X^2} – 1 X$
- $Z_2 = \sqrt[3]{Y + X^3 + \sqrt{Y^2 + 2 X^3 Y}} + \sqrt[3]{Y + X^3 – \sqrt{Y^2 + 2 X^3 Y}} – 2 X$
- $Z_3 = \sqrt{X^2 – v} + \sqrt{2 X^2 + v + \frac{2 X^3}{\sqrt{X^2 – v}}} – 3 X$
- where $v = \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} + \sqrt[3]{X^2 Y – \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}}$
These values are, I discovered, the ONLY positive real roots of the following set of equations:
- $f_0(Z) = Z (Z + 1 X)^0 − 1 Y$
- $f_1(Z) = Z (Z + 2 X)^1 − 1 Y$
- $f_2(Z) = Z (Z + 3 X)^2 − 2 Y$
- $f_3(Z) = Z (Z + 4 X)^3 − 8 Y$
The relationship seems to be that $Z_n$ is the only positive real root of the equation $f_n(Z) = Z (Z + (n + 1) X)^n − 2^{\frac{n^2 – n}{2}} Y$. The coefficient $2^{\frac{n^2 – n}{2}}$ may seem to come out of left field, but it has cropped up repeatedly in the project this is from so it seemed logical to apply it here as well, and it works.
I also have two out of four members of a related but distinct set of values for $Z$:
- $Z_0 = \;???$
- $Z_1 = \sqrt{Y + X^2} – 1 X$
- $Z_2 = \sqrt[3]{Y + \sqrt{Y^2 – \frac{X^6}{27}}} + \sqrt[3]{Y – \sqrt{Y^2 – \frac{X^6}{27}}} – 1 X$
- $Z_3 = \;???$
Both values for $Z_1$ turned out to be the same, but the values for $Z_2$ are different. I am certain that the missing value for $Z_3$ in the second set is different than the known value in the first set, but I don't have enough information to work it out yet. The missing value for $Z_0$ may be the same as the known value in the first set or it may be different; there are multiple possibilities that work, and I don't have enough information to decide which is 'most correct'.
I am hoping that finding a cubic equation whose only positive real root is the second value for $Z_2$ will give me enough information to extrapolate a quartic equation whose only positive real root is the missing value for $Z_3$, and a linear equation whose zero is one of the acceptable values for $Z_0$ to complete the pattern.
Best Answer
While @dxiv has essentially answered the question, we can add some more details for future posts, plus a cameo by the Bring-Jerrard quintic.
I. Deriving formulas
First, if one has an expression of form,
$$Z = -X + \sqrt[m]{a+\sqrt[n]b}+\sqrt[m]{a-\sqrt[n]b}$$
move the bothersome $X$ out of the way by assuming a new variable $\color{blue}{Z= T-X}$ to get the simpler,
$$T = \sqrt[m]{a+\sqrt[n]b}+\sqrt[m]{a-\sqrt[n]b}$$
Once the minimal polynomial for $T$ has been found, it is easy to re-express it in terms of $Z$ using the blue equation.
Second, yes, given one root of any equation, all of its Galois conjugates can be found. The beauty of Galois theory is it unified all known and "future" formulas into a single framework. All equations can be transformed into "depressed" form and, if prime degree $p$ with a solvable Galois group, have predictable formulas. For $p = 2,3,5$, we have
$$z_k = u^{1/2}\,\zeta_2^k$$ $$z_k = u_1^{1/3}\,\zeta_3^k+u_2^{1/3}\,\zeta_3^{2k}$$ $$z_k = u_1^{1/5}\,\zeta_5^k+u_2^{1/5}\,\zeta_5^{2k}+ u_3^{1/5}\,\zeta_5^{3k}+u_4^{1/5}\,\zeta_5^{4k}\\$$
respectively, with roots of unity $\zeta_p = e^{2\pi i/p}$, the $u_i$ as roots of the resolvent equation of degree $p-1$, and $k = (0 \text{ to } p-1),$ implying the equation of degree $p$ has $p$ roots. (Your cubic root was just the case $k=0$.) For example, for $p=3$, then,
$$P(z)=(z-z_0)(z-z_1)(z-z_2) = 0$$
which simplifies to,
$$\color{brown}{P(z)=z^3-3(u_1u_2)^{1/3}z-(u_1+u_2) =0}$$
For cubics, its resolvent is a quadratic, and we have the OP's,
$$u_i = Y \pm \sqrt{Y^2 - \frac{X^6}{27}}$$
So using the $u_i$ on the brown equation, it yields,
\begin{align}P(z) &= z^3-3\Big(\tfrac{X^6}{27}\Big)^{1/3}z-(2Y)=0\\ &= z^3-X^2z-2Y= 0 \end{align}
Let $z = Z+X$ and we recover,
$$Z^3+3XZ^2+2X^2Z−2Y = 0$$ $$Z(Z + 2X)(Z+X) − 2 Y = 0$$
II. Second family and Bring-Jerrard quintic
Per the OP's formula, the first family is,
and so on. The nice thing about this family is that, for any $n$, it has a closed-form solution in terms of generalized hypergeometric functions. Given the more general form,
$$Z (Z + (n+1) X)^n − W =0$$
Let $Z=z^n,$ $X=a,$ $W=(nb)^n,$ thus,
$$\big(z^{n+1} + (n+1)az\big)^n − \big(nb\big)^n =0$$
This is a difference of two $n$th powers and can be factored. We get the trinomial factor,
$$z^{n+1}+(n+1)az-nb = 0\;$$
For example, for $n=4$, yields the Bring-Jerrard quintic,
$$z^5+5az-4b = 0$$
and a root $z$ is,
$$z = \frac{4b}{5a}\,{_4F_3}\Big(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,-\frac{b^4}{a^5}\Big)$$
Then we have $Z = z^n$ by undoing all the substitutions. And so on for any $n$.
III. Second family
The second family seems to be,
for some small integer $a,b,c$, though it hard to ascertain those without more details.