9
$\begingroup$

While trying to answer this question, I got stuck showing that

$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$

The identity is easy to show if you already know the $2+\sqrt{3}$ part; just cube the thing. If you don't know this, however, I am unsure how one would proceed.

That got me thinking ...

If you have some quadratic surd $a+b\sqrt{c}$, where $a$, $b$, and $c$ are integers, and $c$ is not a perfect square, how do you find out if that surd is the cube of some other surd, i.e. how to simplify nested cubic radicals of the form $$\sqrt[3]{a+b\sqrt c}$$

$\endgroup$
3
  • 3
    $\begingroup$ If you take 26+15√3 and multiply it by its conjugate, 26−15√3, then you get its norm, the perfect cube 1. This lets you know that 26+15√3 could be a perfect cube, and that its cube root would have its norm equal to the cube root of 1. Sure enough 2+√3 times 2−√3 is equal to 1. However, this isn't sufficient, since maybe 2+√3 doesn't have a cube root (or if it does, then maybe its cube root has no cube root, etc.). Finding elements of norm 1 is called Pell's Equation and would be one way to check if 2+√3 has another cube root, or if it is finished. $\endgroup$ Jan 4 '11 at 15:00
  • $\begingroup$ IMHO, it's a Good Idea to be familiar with the Pell's equations & the associated continued fraction expansions for the small integers. Eg, if you've played with $x^2-3y^2=1$ there's a good chance that you'd recall that $26^2-3\cdot15^2=1$ $\endgroup$
    – PM 2Ring
    Nov 10 '20 at 16:05
  • $\begingroup$ I am aware that Ramanujan discovered a formula for $\sqrt{a\sqrt[3]b + c\sqrt[3]d}$, namely $$\sqrt{m\sqrt[3]{4m-8n}+n\sqrt[3]{4m+n}}=\frac 13\Big\{\sqrt[3]{(4m+n)^2}+\sqrt[3]{4(m-2n)(4m+n)}-\sqrt[3]{2(m-2n)^2}\Big\}$$ but not sure if it helps $\endgroup$
    – Mr Pie
    Jul 19 at 13:33
5
$\begingroup$

As has been mentioned this is simply solved since it denests already in the field generated by the radicand. Generally this is not true, but there is a general denesting structure theorem that applies. Here's an extract from my Sep 15 post on denesting radicals.

DENESTING STRUCTURE THEOREM$\;\; \;$ Let $\rm\; F \;$ be a real field and $\rm\; F' = F(q_1^{1/d1},\ldots,q_k^{1/dk}) \;$ be a real radical extension of $\rm\; F \;$ of degree $\rm\; n \;$. By $\rm\; B = \{b_0,\ldots, b_{n-1}\}$ denote the standard basis of $\rm\; F' \;$ over $\rm\; F \;$. If $\rm\; r \;$ is in $\rm\; F' \;$ and $\rm\; d \;$ is a positive integer such that $\rm\; r^{1/d} \;$ denests over $\rm\; F \;$ using only real radicals, that is, $\rm\; r^{1/d} \;$ is in $\rm\; F(a_1^{1/t_1},\ldots,a_m^{1/t_m}) \;$ for some positive integers $\rm\; t_i \;$ and positive $\rm\; a_i \in F \;$, then there exists a nonzero $\rm\; q \in F \;$ and a $\rm\; b \in B \;$ such that $\rm\; (q b r)^{1/d} \in F' \;$.

I.e. multiplying the radicand by a $\rm\; q \;$ in the base field $\rm\; F \;$ and a power product $\rm\; b \;=\; q_1^{e_1/d_1}\cdots q_k^{e_k/d_k} \;$ we can normalize any denesting so that it denests in the field defined by the radicand. E.g.

$$ \sqrt{\sqrt[3]5 - \sqrt[3]4} \;\;=\; \frac{1}3 (\sqrt[3]2 + \sqrt[3]{20} - \sqrt[3]{25})$$ normalises to $$ \sqrt{18\ (\sqrt[3]10 - 2)} \;\;=\; 2 + 2\ \sqrt[3]{10} - \sqrt[3]{10}^2 $$

An example with nontrivial $\rm\:b$

$$ \sqrt{12 + 5\ \sqrt 6} \;\;=\; (\sqrt 2 + \sqrt 3)\ 6^{1/4} $$

normalises to

$$ \sqrt{\frac{1}3 \sqrt{6}\: (12 + 5\ \sqrt 6)} \;\;=\; 2 + \sqrt{6} $$

See said post for further details and references.

$\endgroup$
5
  • 3
    $\begingroup$ Could you show how a human could actually do this with non-expert's original example? I love your answer because it gives a name to a long forgotten skill, but the skill itself remains forgotten and unfortunately right now your answer only says to me "for every mysterious denesting, there are even more beautiful, but still mysterious, denestings." $\endgroup$ Jan 4 '11 at 16:03
  • $\begingroup$ +2 for such awesome examples. $\endgroup$ Jan 4 '11 at 16:08
  • $\begingroup$ A nice survey is in Landau's 1994 How to tangle with a nested radical, ams.org/mathscinet-getitem?mr=1270841 dx.doi.org/10.1007/BF03024284 . It also has the Bloemer references and some more crazy examples. It doesn't seem to address the how can humans do it question, but I still found it helpful. $\endgroup$ Jan 4 '11 at 16:28
  • $\begingroup$ @Jack: The point of my post was merely to point out the general denesting structure theorem and give some literature references for the reader. Generally these algorithms are not amenable to hand computation. They may involve nontrivial applications of Galois theory. Further, as I mention in my linked prior post, even very simple looking problems involving radicals have unknown (probably high) complexity. Nonetheless, with luck, knowledge of such techniques can sometimes aid one in hand computations. $\endgroup$ Jan 4 '11 at 17:24
  • $\begingroup$ Your linked post was posted on my birthday lol. But that aside, I have a question out of curiosity. How would we denest something slightly more complicated such as the following? $$\sqrt[3]{5\big(2-\sqrt[5]{27}\big)}$$ which, incidentally, is equal to $\sqrt[5]3-\sqrt[5]9+1$. $\endgroup$
    – Mr Pie
    Jul 19 at 14:23
3
$\begingroup$

Here is a method outlined on page 52 under Algebra: Surds in Carr's Synopsis, the book from which Ramanujan taught himself most of his mathematical skills.

enter image description here

On page 73 in Miscellaneous Equations and Solutions, the author gives an example of how to solve a cubic equation of the form described in the above picture:

enter image description here

And lastly, on page 53, it is claimed that a different and arguably more generalized method can dete[c]t any $\sqrt[n]{A\pm B}$ for odd $n$ such that $A=a\sqrt x$ and $B=b\sqrt y$ for some $\{a,b,x,y\}\subset \mathbb Z^+$ with a supplied example, for which at least one of $x$ and $y$ are square-free.

enter image description here

$\endgroup$
0
2
$\begingroup$

If you write $(d+e \sqrt{f})^3=a+b \sqrt{c}$ and collect terms, you see $c=f$, then $d^3+3de^2=a, 3d^2e+e^3c=b$. For integers, $e$ has to be a factor of $b$, $d$ has to be a factor of $a$ and you can just see if it works pretty easily.

$\endgroup$
4
  • 2
    $\begingroup$ This is pretty good, but d,e don't always have to be integers. Take a=2, b=1, c=5. 2+√5 has a cube root, but I think none with integer coefficients. $\endgroup$ Jan 4 '11 at 15:09
  • $\begingroup$ So, more or less trial and error with the proper divisors of $a$ and $b$? $\endgroup$
    – non-expert
    Jan 4 '11 at 15:11
  • 4
    $\begingroup$ And half-divisors. But yeah, I think this works pretty well. It is like the rational roots "theorem". It sounds dumb "try this big long list" but its much better than "try this infinite list". $\endgroup$ Jan 4 '11 at 15:12
  • $\begingroup$ @Jack Schmidt: I see what you mean. It depends whether you want the cube root in $\mathbb{Z[\sqrt{c}]}$ or $\mathbb{Q[\sqrt{c}]}$. Your point of checking the norm is a good one-it can often tell you when to quit. $\endgroup$ Jan 4 '11 at 15:16
2
$\begingroup$

The following is a technique for solving nested cubic radicals of this form. The technique uses some trigonometric functions by using a polynomial calculator such as PARI, SAGE or an online calculator such as xyiunque.nom.es or WolframAlpha to find roots of a cubic polynomial rather than taking the time to calculate them manually to find solutions. The method is based on either of the following binomial formulas: $(a+b)^3$ or $(a-b)^3$. For question asked above use the $(a+b)^3$ to yield $a^3+3a^2b+3ab^2+b^3$. Set: \begin{align} (1) & a^3+3ab^2=26 \\ (2) & 3a^2b+b^3=15\sqrt{3} \end{align} Solve for b in (1): $b=\sqrt{\frac{26-a^3}{3a}}$. $b^3=\frac{26-a^3}{3a}\sqrt{\frac{26-a^3}{3a}}$ Substitute these into (2) and solve for $a$. \begin{align} (3) & 3a^2\sqrt{\frac{26-a^3}{3a}}+\frac{26-a^3}{3a}\sqrt{\frac{26-a^3}{3a}}=15\sqrt{3}\\ (4) & 9a^3\sqrt{\frac{26-a^3}{3a}}+{26-a^3}\sqrt{\frac{26-a^3}{3a}}=45a\sqrt{3}\\ (5)& \sqrt{\frac{26-a^3}{3a}}(9a^3+26-a^3)=45a\sqrt{3}\\ (6) & \frac{26-a^3}{3a}(64a^6 +416a^3 +676)=6,075a^2\\ (7) & (26-a^3)(64a^6 +416a^3 +676)=18,225a^3\\ (8)& -64a^9+1,248a^6-8,085a^3+17,576=0 \end{align}

There is no need to go through all the above computations for each nested cubic radical we want to denest. We are showing it here only to illustrate the derivation of the standard cubic polynomial formula we will be using. The final coefficients and equation above always take the form $$-64a^9+(48x)a^6+((15x^2)-3(3y)^2)a^3+x^3=0$$ where $x$ and $y$ equal the integer $26$ and the irrational number $15\sqrt3$, under the cubic radical in the question we are addressing. The resulting equation is a cubic equation with $a^3$ as the variable. The equation can be put in monic polynomial form by dividing all the terms by $-64$ if desired. The objective is to ascertain whether the above cubic polynomial has one or more rational roots. If so the nested cubic radical can be denested rather nicely. Rather than go through the computations using trigonometry or other methods to find the roots let's run the equation through a polynomial calculator to find any rational roots. The polynomial $$-64a^9+1,248a^6-8,085a^3+17,576=0$$ has one rational root, $a=2$. Now just substitute $2$ into equation (1) to get $2^3+3(2)b^2=26$ so $b^2=3$ and $b=\pm\sqrt3$. The final result is $a+b$ or $2+\sqrt3$ = $\sqrt[3]{26+15\sqrt{3}}$. The other cubed roots can be computed by multiplying $2+\sqrt3$ by each of the complex cube roots of unity. The other two complex cube roots are $(\frac{-2-\sqrt{3}}{2}$ $+$ $\frac{-2\sqrt{3}-{3}}{2}i)$ and $(\frac{-2-\sqrt{3}}{2}$ $+$ $\frac{2\sqrt{3}+{3}}{2}i)$. The cube root of $\sqrt[3]{26-15\sqrt{3}}$ = $2-\sqrt3$ and its two complex cube roots are $(\frac{-2+\sqrt{3}}{2}$ $+$ $\frac{2\sqrt{3}-{3}}{2}i)$ and $(\frac{-2+\sqrt{3}}{2}$ $+$ $\frac{-2\sqrt{3}+{3}}{2}i)$.

Another Illustration: Let's try another example with the following nested cube root $$\sqrt[3]{3+\frac{10}{3i\sqrt{3}}}$$ So for the purpose of plugging amounts into the standard polynomial derived previously above $x=3$ and $y=\frac{-10i\sqrt{3}}{9}$. The polynomial is derived as follows: \begin{align} (9) & -64a^9+((48(3))a^6+15(3)^2-3(3\frac{-10i\sqrt{3}}{9})^2a^3+27=0 \\ (10) & -64a^9 + 144a^6 + 235a^3 + 27=0\\ (11)& a^9 - \frac{9}{4}a^6 - \frac{235}{64}a^3 - \frac{27}{64}=0\\ (12)& a_1=-1, a_2=-\frac{1}{2},a_3= \frac{3}{2} \end{align} Now that we have the rational roots of the polynomial we can find the three cubic roots of the nested cubic radical by solving the equation $a^3+3ab^2=3$ for $b_1$, $b_2$ and $b_3$ having found the three rational $a$'s. \begin{align} (13) & (-1)^3+3(-1)b_1^2=3\\ (14) & b_1^2=\frac{-4}{3}\\ (15) & b_1=\pm\frac{2i\sqrt{3}}{3}\\ (16) & (-\frac{1}{2})^3+3(-\frac{1}{2})b_2^2=3\\ (17) & b_2^2=\frac{-25}{12}\\ (18) & b_2=\pm\frac{5i\sqrt{3}}{6}\\ (19) & (\frac{3}{2})^3+3(\frac{3}{2})b_3^2=3\\ (20) & b_3^2=\frac{-1}{12}\\ (21)& b_3=\pm\frac{i\sqrt{3}}{6} \end{align} The denested cube roots of $\sqrt[3]{3+\frac{10}{3i\sqrt{3}}}$ are $-1-\frac{2i\sqrt{3}}{3}$; $-\frac{1}{2}+\frac{5i\sqrt{3}}{6}$ and $\frac{3}{2}-\frac{i\sqrt{3}}{6}$. We can check the answers by multiplying any of the solutions by the cube roots of unity $-\frac{-1}{2}-\frac{i\sqrt{3}}{2}$ and $-\frac{-1}{2}+\frac{i\sqrt{3}}{2}$ to get each of the other two solutions. For free we also get the three solutions to the nested radical $\sqrt[3]{3-\frac{10}{3i\sqrt{3}}}$ which are $-1+\frac{2i\sqrt{3}}{3}$; $-\frac{1}{2}-\frac{5i\sqrt{3}}{6}$ and $\frac{3}{2}+\frac{i\sqrt{3}}{6}$. One of the drawbacks to this method is that the coefficients in the polynomial can get quite large which can make working with it a bit unwieldy. If the polynomial roots are not rational the solutions to the cube roots we are looking for will not look as nice but they will be correct. For example try to show that $$\sqrt[3]{45+\frac{\sqrt{40\sqrt{3}-1}(40\sqrt{3}+8)}{8}} = \frac{\sqrt{3}+\sqrt{40\sqrt{3}-1}}{2}$$ by deriving the following cubic polynomial and finding a fairly familiar looking irrational root of the equation $$a^9 - \frac{135}{4}a^6 + \frac{81,810\sqrt{3}-27}{64}a^3 - \frac{91,125}{64}=0$$ You can find that root on WolframAlpha in symbolic notation or for that matter compute the cube root of the nested radical in symbolic notation on WolframAlpha. I do not think all nested cubic radicals can be denested with radicals in the solution. Try to show that $$\frac{\sqrt{3}+\sqrt{40\sqrt{3}-1}}{2} = \frac{2\sqrt{3}+\sqrt{-2+2i\sqrt{4,799}}+\sqrt{-2-2i\sqrt{4,799}}}{4}$$ by using the denesting method described in Stack Exchange answer to question number 220818 to denest (in this case renest) the nested radical on the left hand side of the equality.

$\endgroup$
2
$\begingroup$

Apply the known denesting formula $$\sqrt[3]{a+b \sqrt c}=\frac12\sqrt[3]{3bt-a}\left(1+\frac1t \sqrt c\right)$$ where $t$ satisfies $t^3-\frac{3a}bt^2+3c t-\frac{ac}b=0$. Take the example $\sqrt[3]{26+15\sqrt{3}}$ in question and solve $$t^3-\frac{26}5t^2+9t-\frac{26}5=\frac15(t-2)(5t^2-16t+13)=0$$ which yields $t=2$ and the denestation $$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$


Other examples are listed below, along with their resolvent equations \begin{align} \sqrt[3]{7+5\sqrt{2}}= 1+\sqrt{2}&\>\>\>\>\>\>\> t^3-\frac{21}5t^2+6t-\frac{14}5 =0,\>\>\>t=1\\ \sqrt[3]{90-34\sqrt{7}}= 3-\sqrt{7}&\>\>\>\>\>\>\> t^3+\frac{135}{17} t^2+21 t+\frac{315}{17}=0 ,\>\>\>t=3 \\ \sqrt[3]{\frac{99}2+\frac{59}2\sqrt{\frac52}}= 3+\sqrt{\frac52}&\>\>\>\>\>\>\> t^3-\frac{297}{59} t^2+\frac{15}2 t-\frac{495}{118}=0 ,\>\>\>t=3 \\ \sqrt[3]{25+10\sqrt{5}}= \frac52+\frac12\sqrt{5}&\>\>\>\>\>\>\> t^3-\frac{15}{2} t^2+15 t-\frac{25}{2}=0 ,\>\>\>t=5\\ \sqrt[3]{70-22\sqrt{7}}= \sqrt[3]{49}\left(1-\frac{\sqrt{7}}7\right)&\>\>\>\>\>\>\> t^3+\frac{105}{11} t^2+21 t+\frac{245}{11}=0 ,\>\>\>t=7\\ \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.