5

Unclear how valuable this posting is. It really should be limited to specifying that the goal is to denest one level of the radicals, in an expression like

$$\left[c + d\sqrt{D}\right]^{1/3} + \left[c - d\sqrt{D}\right]^{1/3} ~c,d,D \in \Bbb{Z}, ~D~ \text{is square free}.$$ As KCd indicated in a comment, following his answer, I totally overlooked that the rational root theorem is decisive for finding a rational value for the variable $a$.

So, the (remaining) problem is : what happens if

$$\left(a + b\sqrt{D}\right)^3 = c + d\sqrt{D},$$

where $a$ is irrational?

$\underline{\text{The Problem}}$

I am looking for an Algebraic derivation that

$$\left[ ~a + b\sqrt{57} ~\right]^3 = \left[ ~540 + 84\sqrt{57} ~\right] ~: ~a,b \in \Bbb{R}$$

may be solved by $~(a,b) = (3,1).$

$\underline{\text{My Background}}$

Some years ago, I survived self-studying :

  • Calculus, Vol 1, 2nd Ed. (Tom Apostol, 1966)

  • Through chapter 10, which includes Quadratic Reciprocity Law, of
    Elementary Number Theory (Uspensky and Heaslett, 1938)

  • Chapters 1 and 2 only of
    An Introduction to Complex Function Theory (Bruce Palka, 1991).

$\underline{\text{Problem Background}}$

I noticed a youtube problem: $~\displaystyle f(x) = x^3 + x - \frac{5}{8} = 0.$

Since I trial/error saw that $~f(1/2) = 0,$ I was able to use
polynomial long division to determine that the roots of $~f(x) = 0~$ are

$\displaystyle \left( ~\frac{1}{2}, \frac{-1 \pm i\sqrt{19}}{4} ~\right).$

As an exercise, I decided to practice using Cardano's Method against the equation:

$$x^3 + x - \frac{5}{8} = 0.$$

Setting

$$S + T = x, ~3ST = -1 \implies x^3 = S^3 + T^3 = S^3 + \left[\frac{-1}{3S}\right]^3 \implies $$

$$\left[S^3\right]^2 - \frac{5}{8}\left[S^3\right] - \frac{1}{27} = 0 \implies $$

$$S^3 = \frac{1}{2} ~\left[ ~\frac{5}{8} \pm \frac{7}{72}\sqrt{57} ~\right]$$

$$= \frac{1}{\left(12\right)^3} ~\left[540 \pm 84\sqrt{57}\right].$$

This implies that the equation

$$x^3 + x - \frac{5}{8} = 0$$

has the real root

$$\frac{1}{12} ~\left( ~\left[540 + 84\sqrt{57}\right]^{(1/3)} ~+~ ~\left[540 - 84\sqrt{57}\right]^{(1/3)} ~\right). \tag1 $$

$\underline{\text{My Initial Work}}$

In order to simplify the expression in (1) above, I noted that

$$\left[a + b\sqrt{57}\right]^3 = \left[a^3 + 171ab^2\right] + \sqrt{57} ~\left[3a^2b + 57b^3\right].$$

So, I have the following two (non-linear) equations in two unknowns:

  • Equation-1 : $~\displaystyle a^3 + 171ab^2 = 540.$

  • Equation-2 : $~\displaystyle 3a^2b + 57b^3 = 84.$

Since I couldn't find an obvious line of attack to derive the $~(a,b) = (3,1)~$ solution to the above two equations, I took the preliminary step of verifying the solution. I used a somewhat convoluted method.

I reasoned that since the only real root of $~f(x) = x^3 + x - \frac{5}{8} = 0~$ is $~x = \frac{1}{2} = \left[\frac{3}{12} + \frac{3}{12}\right],~$ I must have that $~a = 3.~$ I was then able to verify that $~(a,b) = (3,1)~$ satisfied both of Equation-1 and Equation-2, above.

$\underline{\text{My Subsequent Work}}$

Since the derivation process involves not knowing any of the actual roots to $~f(x) = 0,~$ the $~(a,b) = (3,1)~$ guesswork does not represent an analytical means of attack.

One approach is to substitute one value for another.

This leads to (for example)

$$3a^2 \left[ ~\frac{540 - a^3}{171a} ~\right]^{(1/2)} + 57\left[ ~\frac{540 - a^3}{171a} ~\right]^{(3/2)} = 84. \tag2 $$

Edit
The above expression does not represent a Gauss function.

My only other try is to try to use elementary Complex Analysis, by noting that

$$\left[a + ib\sqrt{57}\right]^3 = \left[a^3 - 171ab^2\right] + i\sqrt{57} ~\left[3a^2b - 57b^3\right].$$

If I could (somehow) obtain an appropriate expression of

$$(a + ib)^3 = \left[a^3 + 171ab^2\right] + i\sqrt{57} ~\left[3a^2b + 57b^3\right],$$

then, I could convert the RHS above into $~re^{i\theta},~$ thereby simplifying the cube root to

$$r^{1/3}e^{i[\theta + 2k\pi]/3} ~: ~k \in \{0,1,2\}.$$

However, I see no way of pursuing this last approach.

user2661923
  • 42,303
  • 3
  • 21
  • 46

4 Answers4

4

To do this in a non-ugly way, use the norm map $N(x+y\sqrt{57}) = x^2 - 57y^2$, which is multiplicative from $\mathbf Z[\sqrt{57}]$ to $\mathbf Z$.

Suppose $(a + b\sqrt{57})^3 = 540 + 84\sqrt{57}$ for some integers $a$ and $b$. Take the norm of both sides: $$ (a^2 - 57b^2)^3 = 540^2 - 57 \cdot 84^2 = -110592= (-48)^3, $$ which is equivalent to $a^2 - 57b^2 = -48$. Since $57$ and $48$ are both divisible by $3$, $a$ must be divisible by $3$, so write $a = 3c$. Plug that in and divide through by $3$ to get $$ 3c^2 - 19b^2 = -16, $$ or equivalently $3c^2 = 19b^2 - 16$. An example of an integral solution to that is obvious: $c = 1$ and $b = 1$, so we try $a = 3c = 3$ and $b = 1$ in the original equation and it works.

KCd
  • 55,662
  • Will probably eventually accept your answer, since I suspect that no one will come closer. I envision two problems, attempting to use your method for general situations of this type: [1] $a,b$ do not necessarily have to be integers. [2] When attacking $$\left(a + b\sqrt{D}\right)^3 = c + d\sqrt{D} ~: ~D~ \text{is square free},$$ $c^2 - Dd^2~$ might not be a perfect cube. – user2661923 Jan 19 '23 at 02:35
  • I thought the purpose was to explain how to search for an integral solution. If you want integers $a, b, c, d$ such that $(a + b\sqrt{D})^3 = c+d\sqrt{D}$ then $(a^2-Db^2)^3 = c^2 - Dd^2$, so $c^2 - Dd^2$ is a cube in $\mathbf Z$. (We need $D$ not a square, like $D = 12$; it need not be squarefree.) And similarly, if there is a solution where $a, b, c, d$ are rational then $c^2 - Dd^2$ is a rational cube. Why are you interested in such equations without a constraint like integral or rational unknowns? For unknowns in $\mathbf R$ it is trivial (let $b = 0$ and $a = \sqrt[3]{c+d\sqrt{D}}$). – KCd Jan 19 '23 at 02:43
  • No, see the very start of my posting. In the generic problem, when you employ Cardano's method, you will frequently get an expression like $$\left[c + d\sqrt{D}\right]^{1/3} + \left[c - d\sqrt{D}\right]^{1/3}.$$ I need to simplify this to $$\left(a + b\sqrt{D}\right) + \left(a - b\sqrt{D}\right) = 2a ~: ~a \in \Bbb{R}.$$ – user2661923 Jan 19 '23 at 02:47
  • Actually, I am interested in rational solutions, if one is feasible. The difficulty with the real solution of $$a_1 = \left[c + d\sqrt{D}\right]^{1/3}, a_2 = \left[c - d\sqrt{D}\right]^{1/3},$$ is that the nested radical has not been removed. I guess that I should have emphasized that the purpose is to simplify by denesting the radicals. – user2661923 Jan 19 '23 at 02:52
  • I am interested in your comment that if rational satisfying $a,b$ are feasible that then $c^2 - Dd^2$ is a rational cube. This makes sense except: is it plausible that you might know that $c^2 - Dd^2$ is a rational cube, of the form $~\dfrac{P^3}{Q^3},~$ without knowing the value of $Q$? In effect, with integers, you know that the denominator is $1$. If $Q$ is unknown, can your approach still be made to work? – user2661923 Jan 19 '23 at 02:56
  • So are you starting with rational $c, d$ and you want to find rational $a$ and $b$? Usually you can't: if $c+d\sqrt{D}$ is going to be $(a+b\sqrt{D})^3$ for some rational $a$ and $b$, then necessarily $c^2 - Dd^2 = (a^2 - Db^2)^3$, so $c^2 - Dd^2$ must be a rational cube. In actual examples, like the one you gave, this can be checked because the right side is given to us (are we not allowed to compute with it?). Suppose it is $r^3$. Then $a^2 - Db^2 = r$, which is a generalized Pell equation ($D$ and $r$ known, $a$ and $b$ unknown). Approach this with Cardano's method is simply a bad idea. – KCd Jan 19 '23 at 02:59
  • Re my previous comment, rather than the premise that you know that $c^2 - Dd^2$ is a rational cube, you could merely take that as a hypothesis, thus exploring whether rational satisfying $a,b$ exist. Again, this hinges on where it is okay that $Q$ is unknown. – user2661923 Jan 19 '23 at 02:59
  • Per your last comment, that makes a lot of sense. "Approach this with Cardano's method is simply a bad idea.": How else can you attack a cubic equation besides the rational root theorem, or Cardano's method? – user2661923 Jan 19 '23 at 03:02
  • The problem you posed is not just any old cubic equation: it has extra structure that I used. Now if you want to decide if an arbitrary cubic polynomial with rational coefficients has a rational root, you've changed the problem. Methods that work on a special case need not work in the general case. For a general cubic with rational coefficients, use the rational roots theorem. Cardano is simply awful. – KCd Jan 19 '23 at 03:05
  • Good point. I got so caught up in looking for a generic denesting algorithm, that I totally overlooked that the rational root theorem itself is decisive. "Cardano is simply awful" : Agreed. – user2661923 Jan 19 '23 at 03:10
1

Apply the cubic-root denesting formula $$\sqrt[3]{a+\sqrt b}=\frac{\sqrt[3]{3p-a}}{2p}\left(p+\sqrt b\right)$$ where $p$ is the root of $p^3-{3a}p^2+3b p-{ab}=0$. For $\sqrt[3]{540 + 84\sqrt{57}}$, solve $$p^3-3\cdot 540 \ p^2+3\cdot 57\cdot 84^2\ p-540\cdot 57\cdot 84^2=0$$ to get $p=252$. Plug into the denesting formula above to arrive at $$\sqrt[3]{540 + 84\sqrt{57}}=3+\sqrt{57}$$

Quanto
  • 120,125
  • +1 : +Accept to this answer, which is simply a special case of the more general answer that you gave to this question. In fact, I liked your general answer so much, that whenever the issue comes up on MathSE, I always refer people to your general formula. Unfortunately, the only downside is that you then have to solve another cubic, perhaps again using Cardano's method, so it is theoretically possible to end up chasing your tail. ...see next comment – user2661923 Jan 26 '25 at 17:17
  • My personal experience however, is that in the 4 or 5 times that I've applied your general approach, using Cardano's method for the derived cubic, I have never had to chase my tail (for one reason or another). Further, none of the other answers that I have seen provide more elegant more feasible (purely analytical/algebraic) approaches. So, I suspect that your approach is as least as good as any other approach. – user2661923 Jan 26 '25 at 17:22
0

$(a+b\sqrt{D})^3 = (3ab^2D+a^3) + (b^3D+3a^2b)\sqrt{D} = x + y\sqrt{D}$

$x = (3ab^2D+a^3) \quad → \displaystyle b^2 = \frac{x-a^3}{3aD}$

$ y = b\,(Db^2+3a^2)$

$\displaystyle y^2 = \left(\frac{x-a^3}{3aD}\right)\left( \left(\frac{x-a^3}{3a}\right) + 3a^2 \right)^2$

Let $A=a^3$, and multiply both side by $27DA$, we have:

$27Dy^2A = (x-A)(x+8A)^2$

$64A^3 - 48x\,A^2 + (27Dy^2-15\,x^2)A - x^3 = 0$

Derive similarly, we also have a cubic for $B = b^3$

$27B\,x^2 = (y-DB)\big(y + 8DB \big)^2$

$64D^3\,B^3 - 48D^2y\,B^2 + (-15Dy^2+27x^2)B - y^3 = 0$

albert chan
  • 2,204
  • I am looking for an analytical solution to the general problem of $$\left(x + y\sqrt{D}\right)^3 = c + d\sqrt{D} ~: D~ \text{is square free}.$$ Computer software is off-limits. – user2661923 Jan 19 '23 at 02:39
  • That should be easy enough. I will update later ... – albert chan Jan 19 '23 at 02:40
  • I am going to have to regard any solution that requires you to solve a separate cubic equation as also off-limits. Otherwise, for the general problem, you will often be chasing your tail, going from one problematic cubic equation to another. – user2661923 Jan 19 '23 at 02:42
  • We can't solve a cubic to get (a,b)? What if they are irrational? Your question clearly stated (a,b) is real ... – albert chan Jan 19 '23 at 02:50
  • $(a,b)$ is real. Real numbers can be irrational. The goal is to de-nest one of the nested radicals, in an expression like $$\left[c + d\sqrt{D}\right]^{(1/3)} - \left[c - d\sqrt{D}\right]^{(1/3)}.$$ As for using a cubic equation to solve for $a,b$, a cubic equation is what caused the nested radical in the first place. What happens when the cubic equation that you solve to compute $(a,b)$ itself involves nested radicals? – user2661923 Jan 19 '23 at 03:06
  • I do not know a general algorithm to de-nest radicals. It might not always be possible. – albert chan Jan 19 '23 at 03:57
0

My previous answer assumed $a,b \in \Bbb{R}$, solved algebraically for $\sqrt[3] {x+y\sqrt{D}} = a+b\sqrt{D}$
$(a^3,b^3)$ are real solutions of cubic equations, which may be messy.

If the goal is to simplify, and $a,b \in \Bbb{Q}$, we can do this numerically.

$\sqrt[3]{A ± \sqrt{R}} = a ± \sqrt{r} \quad ⇒ \quad a = \large\frac{\sqrt[3]{A+\sqrt{R}} \;+\; \sqrt[3]{A-\sqrt{R}}}{2}$

I discussed this very issue in HP Forum: HP50g simplifying a root
The link has Lua code to do simplifying (with confirmation) automatically.

lua> x, y, D = 540, 84, 57
lua> (cbrt(x + y*sqrt(D)) + cbrt(x - y*sqrt(D))) / 2
3.0000000000000004

lua> simp_cbrt4(x, y, D)
3       1       57


lua> simp_cbrt4(81, 30, -3) -- integer argument might produce halves
4.5     0.5     -3

$\displaystyle \sqrt[3]{540+84\sqrt{57}} = 3 + \sqrt{57}$

$\displaystyle \sqrt[3]{81+30\sqrt{-3}} = \frac{9+\sqrt{-3}}{2}$

albert chan
  • 2,204