How to simplify nested cubic radicals $\sqrt[3]{a+b\sqrt c}$

Question

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 $$\sqrt[3]{a+b\sqrt c}$$

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

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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}

Answer 5

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.

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:

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.