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This is a review problem for an introductory Galois theory course.

Rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$.

There could be many ways to do this, but it's implicit that we use field theory for this problem. I tried looking at the solutions to this question, this question and this question, but they haven't helped me so far.

What seems clear is to replace $x = \sqrt[3]{5}$ and then our expression becomes $\frac{1}{1+x-x^2}$.

Now, since $x^3-5$ is the minimal polynomial of $\sqrt[3]{5}$ in $\mathbb Q[x]$, I think I want to find the inverse of $1+x-x^2$ in the quotient ring $\mathbb Q[x]/(x^3-5)$, but I'm not sure how.

Any hints or suggestions would be appreciated.

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    $\begingroup$ Use the Euclidean algorithm to find a way to express $1$ as a multiple of $1+x-x^2$ plus a multiple of $x^3-5$. $\endgroup$
    – Arturo Magidin
    Commented May 10, 2023 at 3:38
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    $\begingroup$ Alternatively, you can multiply the numerator and denominator by $(1+\omega\sqrt[3]{5}-\omega^2\sqrt[3]{25})(1+\omega^2\sqrt[3]5-\omega\sqrt[3]{25}),$ where $\omega$ is a primitive cube root of $1.$ Using that $\omega+\omega^2=-1.$ But that is harder than the Euclidean algorithm technique. $\endgroup$
    – Thomas Andrews
    Commented May 10, 2023 at 3:51
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    $\begingroup$ For yet another alternative, you could write $\frac{1}{1+\sqrt[3]{5}-\sqrt[3]{25}}=a+b\sqrt[3]{5}+c\sqrt[3]{25}$ then cross-multiply and identify coefficients to find rational $a,b,c$ that satisfy the equation. But the $\gcd$ hint in the first comment is the better way to go. $\endgroup$
    – dxiv
    Commented May 10, 2023 at 4:38
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    $\begingroup$ @User You should clarify in the description of the bounty what you are looking for which is not already covered in the questions linked by the OP, the previous comments above, or this other question. $\endgroup$
    – dxiv
    Commented May 13, 2023 at 0:12
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    $\begingroup$ @lonestudent It does not require Galois theory. There are several different ways to solve it, as mentioned in my previous comment. That's why it's not clear to me what "canonical answer is supposed to mean in the description of the bounty. $\endgroup$
    – dxiv
    Commented May 13, 2023 at 0:46

9 Answers 9

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Set $z=\sqrt[3]{5}$ and solve with the help of $z^3=5$ \begin{align*} (az^2+bz+c)\cdot (-z^2+z+1)&=1\\[6pt] -az^4+(a-b)z^3+(a+b-c)z^2+(b+c)z+c&=1\\[6pt] (a+b-c)z^2+(b+c-5a)z+(5a-b5+c)&=1\\[6pt] c=a+b\; , \;b=2a\; , \;1=5a-10a+a+2a=-2a&\\[6pt] a=-0.5\, , \,b=-1\, , \,c=-1.5& \end{align*} to get $$ \dfrac{1}{1+\sqrt[3]{5}-\sqrt[3]{25}}=\dfrac{1}{-z^2+z+1}=-\dfrac{1}{2}z^2-z-\dfrac{3}{2}=-\dfrac{1}{2}\sqrt[3]{25}-\sqrt[3]{5}-\dfrac{3}{2} $$

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    $\begingroup$ This method of undetermined coef's is usually much more work by hand than using the extended Euclidean algorithm (which here uses only two simple poly divisions). That's true in this case, but may not be obvious since most of the work is omitted above, such as computing the product polynomial and solving the system of equations. $\endgroup$
    – Bill Dubuque
    Commented May 18, 2023 at 18:31
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Now, since $x^3-5$ is the minimal polynomial of $\sqrt[3]{5}$ in $\mathbb Q[x]$, I think I want to find the inverse of $1+x-x^2$ in the quotient ring $\mathbb Q[x]/(x^3-5)$, but I'm not sure how.

While the posted answers are all worthy, as well as those under the related How to rationalize denominator?, the following is an attempt to stay closer to the above formulation of the question.

Idea is to  (1)  determine a cubic equation that $1+x-x^2$ satisfies in $\mathbb Q[x]/(x^3-5)$, then  (2)  multiply it by $(1+x-x^2)^{-1}$ and calculate the inverse.

(1)   Let $\,g = 1 + x - x^2\,$, then: $$ \begin{align} &g^3 - 3 g^2 + 18 g + 4 = -(x^3 - 5) (x^3 - 3 x^2 + 3 x + 4) \tag{1} \\ \implies\quad\;\; &g^3 - 3 g^2 + 18 g + 4 \equiv 0 \pmod{x^3 - 5} \tag{2} \end{align} $$ Polynomial $(1)$ was not magically pulled out of a hat. A step-by-step derivation using only elementary means is given for a more general case in my answer here to the question Is $a+5^{1/3}b+5^{2/3}c$ a root of any cubic polynomial in $\mathbb{Q}$? (this one here corresponds to $\lambda = \mu = 0, \nu = 5, a=b=1, c=-1$ in my other post).

(2)   $x^3-5\,$ is irreducible over $\mathbb Q$, so it is coprime to all rational polynomials of degree $\lt 3$. It follows that $\,g=1+x-x^2\,$ is invertible in $\mathbb Q[x]/(x^3-5)$, then multiplying $(2)$ by $g^{-1}$ and solving for $g^{-1}\,$: $$ \begin{align} &&4g^{-1} &\equiv -g^2 + 3 g - 18 \\ &&&\equiv -(1+x-x^2)^2 + 3 (1+x-x^2) - 18 \\ &&&\equiv -x^4 + 2 x^3 - 2 x^2 + x - 16 \\ &&&\equiv -(x^3 - 5)(x - 2) -2 (x^2 + 2 x + 3) \\ &&&\equiv -2 (x^2 + 2 x + 3) \pmod{x^3 - 5} \\ \implies\;\; &&(1+x-x^2)^{-1}\;\; &\equiv - \frac{1}{2}(3 + 2x + x^2) \pmod{x^3 - 5} \end{align} $$

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    $\begingroup$ @lonestudent Yes, any algebraic expression in algebraic numbers is algebraic itself, and can therefore be inverted/rationalized using some of the same techniques. $\endgroup$
    – dxiv
    Commented May 14, 2023 at 4:55
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    $\begingroup$ This method is also already presented in many prior answers, e.g. see "generally" here where I present it from a more general perspective. $\endgroup$
    – Bill Dubuque
    Commented May 14, 2023 at 16:35
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    $\begingroup$ I like the other answers as well, but I'm accepting this answer since it is the method I originally had in mind. $\endgroup$
    – pyridoxal_trigeminus
    Commented May 14, 2023 at 17:39
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    $\begingroup$ @pyridoxal That makes more sense. Note that you can do it more Galois theoretically by multiplying by conjugates to get the norm, which is essentially the same as using the constant coef of some poly having it as root (as above) - see my answer linked in above comment. But for hand computation the forward Ext. Euclidean Algorithm is almost always (much) simpler, so I recommend that you learn that if you need to do such computations (if anything is unclear I can elaborate). $\endgroup$
    – Bill Dubuque
    Commented May 14, 2023 at 18:00
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    $\begingroup$ @lone in your method (revision $785$ here) we can use elimination (e.g. resultants) to compute the sought multiple (norm), e.g. $\,{\rm resultant}(1+x−x^2,\, y−x^3)=−y^2+4y+1\,$. Resultants are a common way of computing norms (= product of conjugates), e.g. see here $\ \ $ $\endgroup$
    – Bill Dubuque
    Commented May 18, 2023 at 20:12
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We use $\color{red}{\text{Extended Euclidean Algorithm}}$ (for polynomials), which is a general method to deal with this type of rationalization problems, without introducing any tricks.

Let $P=x^3-5, Q=-x^2+x+1$, our goal is to express the constant term as the combination of $P$ and $Q$, such as

$$1=a(x)P+b(x)Q\Longleftrightarrow \frac{1}{Q}=\frac{a(x)P}{Q}+b(x)$$

If we set $x=\sqrt[3]{5}$, then term $P=0$, we get

$$\frac{1}{Q}=b(x)$$

So the term $b(x)$ will be the answer to this problem.

$$===============\text{START}===============$$

Do long division for $\frac{P}{Q}$, and we get

$$P=(-1-x)Q+\color{red}{R}\tag{1}$$

where $\color{red}{R}=2x-4$, do long division for $\frac{Q}{\color{red}{R}}$, and we get

$$Q=\left(-\frac{1}2x-\frac{1}2\right)\color{red}{R}-1\tag{2}$$

Plug eq.(1) into eq.(2) to eliminate $\color{red}{R}$, we get

$$1=\left(-\frac{1}2x-\frac{1}2\right)P+\left(-\frac{1}2x^2-x-\frac{3}2\right)Q$$

Set $x=\sqrt[3]{5}$, then term $P=0$, we get

$$\boxed{\frac{1}Q=\frac{1}{1+\sqrt[3]{5}-\sqrt[3]{25}}=-\frac{1}2\sqrt[3]{25}-\sqrt[3]{5}-\frac{3}2}$$

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    $\begingroup$ As usual, it's much easier to use the forward extended Euclidean algorithm - see my answer. $\endgroup$
    – Bill Dubuque
    Commented May 14, 2023 at 6:45
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My goal will be to simply derive expressions in the denominator in terms of $\thinspace x^{3n}\thinspace . $

Indeed, $\thinspace x=\sqrt [3]{5}\thinspace $ leads to the fraction $\thinspace \frac{1}{1+x-x^2}\thinspace $ or $\thinspace -\frac{1}{x^2-x-1}\thinspace .$

Then, by applying the formula $\thinspace a^3\pm b^3=\left(a\pm b\right)\left(a^2\mp ab+b^2\right)\thinspace $ twice, you have :

$$ \begin{align}-\frac{1}{\left(x^2-x+1\right)-2}&=-\frac {x+1}{x^3+1-2(x+1)}\\ &=-\frac {x+1}{\left(x^3-1\right)-2x}\\ &=-\tfrac {(x+1)\left(\left(x^3-1\right)^2+2x\left(x^3-1\right)+4x^2\right)}{\left(x^3-1\right)^3-8x^3} \thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$

Putting $\thinspace x=\sqrt [3]{5}\thinspace $, then the original fraction becomes :

$$ \begin{align}\frac{1}{1+\sqrt[3]{5}-\sqrt[3]{25}}&=-\frac 12 \big(3+2\sqrt [3]{5}+\sqrt [3]{25}\big)\end{align} $$

which completes the answer .

In the context of Bill Dubuque's explanatory comments ( 1 and 2 ) , the following method has also been added as a minor addition to the answer .

Let $\thinspace P(x)\thinspace $ be a polynomial. We want to multiply the denominator by a polynomial $\thinspace P(x)\thinspace $ such that the denominator can only be written in terms of $\thinspace x^{3n}$ :

$$ \begin{align}\frac {1}{1+x-x^2}&=-\frac {P(x)}{P(x)(x^2-x-1)}\end{align} $$

The construction of the polynomial $P(x)$ can be done as follows :

Let $\thinspace x=z\thinspace $ be one of the roots of $\thinspace x^2-x-1\thinspace .$ This implies that, $\thinspace z^2=z+1\thinspace $ and this leads to the following :

$$ \begin{align}z^6&=z^3+1+3z(z+1)\\ &=z^3+1+3z^3\\ &=4z^3+1\end{align} $$

Then, by quickly applying synthetic division, we have :

$$ \begin{align}&P(z)(z^2-z-1)=z^6-4z^3-1\\ \implies &P(z)=\frac {z^6-4z^3-1}{z^2-z-1}\\ \implies &P(z)=z^4+z^3+2z^2-z+1\end{align} $$

Putting $\thinspace x=\sqrt [3]{5}\thinspace $ and $\thinspace x^3=5\thinspace $, then we reach the following conclusion :

$$ \begin{align}\frac {1}{1+x-x^2}&=-\frac {x\left(x^3\right)+x^3+2x^2-x+1}{x^6-4x^3-1}\\ &=-\frac {2x^2+4x+6}{4}\\ &=-\frac 12 \big(3+2\sqrt [3]{5}+\sqrt [3]{25}\big)\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$

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    $\begingroup$ Neat calculation shortcut (+1). In the general case $\dfrac{1}{a_0+a_1x+a_2x^2}$ you would multiply with $(a_2 x - a_1)$ at the first step, then the rest works similarly, which gives another answer to How to rationalize denominator? $\endgroup$
    – dxiv
    Commented May 13, 2023 at 2:25
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    $\begingroup$ @dxiv Thank you for mention the related link and reviewing the answer. I was very confused by the Galois theory in the tags of the question. (also, I have no idea how rationalizing the denominator relates to Galois theory). I stated in the first version of my answer that the computation lacks Galois theory and frankly, I planned to delete the answer. But, based on your comment above, I left it as it is . $\endgroup$
    – lone student
    Commented May 13, 2023 at 3:55
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    $\begingroup$ @dxiv This is a special case of Gauss's inversion algorithm, i.e. it's the same as my answer except instead of modding the prior remainder by the current one we instead always mod the modulus $p=x^3-5$ by the current one. It corresponds to computing $(p,f) = 1$ for prime $p$ by using only Euclidean descent steps of form $(p,g) = (p, p\bmod g)$. This answer is equivalent to the special case $f$ is quadratic (so descent takes at most $2$ steps). If there is interest I can give further details $\endgroup$
    – Bill Dubuque
    Commented May 15, 2023 at 16:53
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    $\begingroup$ @dxiv I appended a note to my answer showing in detail how the above answer is a special case of Gauss's inversion algorithm. At lone: it would be much clearer if you reduce the fractions using $x^3\equiv 5$ immediately (vs. at end). Then your first reduced fraction is $\,-(x+1)/(4-2x)\,$ and the final arises by scaling top & bottom by $(4^3-(2x)^3)/(4-2x) = 4^2+4(2x)+(2x)^2,\,$ as explained in my answer. $\endgroup$
    – Bill Dubuque
    Commented May 15, 2023 at 21:05
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    $\begingroup$ The method you mention of computing the norm by elimination (e.g. by resultant) is so infrequently explicitly presented here that I think it is worth adding here to help it get better exposure $\endgroup$
    – Bill Dubuque
    Commented May 18, 2023 at 21:31
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Below we use a few twists on the extended Euclidean algorithm and we explain in detail how lone student's slick answer is a special case of one of these (Gauss's modular inversion algorithm).

Let $\,x = \sqrt[3]5,\,$ so $\,\color{#90f}{x^3 = 5},\,$ so $\,(\color{c00}{x\!-\!a})(\underbrace{x^2\!+\!ax\!+\!a^2}_{\textstyle \color{#b70}{g_a}})=\!\color{#90f}{\underbrace{x^3}_{5}}\!\!-a^3 = \begin{cases}\ \ \ \color{#0aa}{6,\,\ a=-1}\\[.1em] \color{#b70}{-3,\,\ a\ =\ 2}\end{cases}$

${\rm so} \ \ \dfrac1{(\color{#0aa}{x^2\!-\!x\!+\!1})\!-\!2}\ {\dfrac{x\!+\!1}{\color{#0aa}{x\!+\!1}}} =\dfrac{x\!+\!1}{\color{#0aa}6\!-\!2(x\!+\!1)} =\dfrac{\color{0a0}{3\!+\!\color{#b70}{x\!-\!2}\!}}{-2\,(\color{#b70}{x\!-\!2})}\ \color{#b70}{\dfrac{g_2}{g_2}}= \dfrac{3g_2\!\color{#b70}{-\!3}}{-2(\color{#b70}{-3})}=\color{#c00}{\dfrac{g_2\!-\!1}{2}}$

That's the essence of lone student's proof. We show below how it is related to the more general algorithm for modular inversion by the extended Euclidean algorithm (in particular, we show how the above method is a special case of Gauss's algorithm for modular inversion).

We compute $\,\color{#c00}{f\equiv (x^2\!-\!x\!-\!1)^{-1}}\pmod{x^3\!-5}\,$ via Forward Extended Euclidean Algorithm

fractionally: $\ \ \ \ \dfrac{0}{\color{#90f}{x^3\!-5}}\overset{\large\frown}\equiv\dfrac{1}{x^2\!-\!x\!-\!1}\overset{\large\frown}\equiv\,\dfrac{\!\!-x\!-\!1}{\color{#0a0}{2x\!-\!4}}\overset{\large\frown}\equiv\color{#c00}{\dfrac{x^2\!+\!2x\!+\!3}2=\dfrac{g_2\!-\!1}2},\ $ in equation form:

$\!\!\!\begin{array}{rrl}\bmod\ \color{90f}{x^3-5}\!:\qquad\ \ [\![1]\!] &\color{#90f}{(x^3\!-5)}f&\!\!\!\equiv\, 0\\ [\![2]\!]&\!\! (x^2\!-x\!-\!1)f&\!\!\!\equiv\, 1;\ \ \ \ \ \ \ \ \ \ \color{#90f}{x^3\!-\!5} = \smash[t]{\overbrace{\color{#0aa}{(x\!+\!1)}}^{\rm quotient}(x^2\!-x\!-\!1) +\!\! \overbrace{\color{#0a0}{2x\!-\!4}}^{\rm remainder}}\to\\ [\![1]\!]\!-\!\color{#0aa}{(x\!+\!1)}[\![2]\!] = [\![3]\!] &\color{#0a0}{(2x\!-\!4)}f&\!\!\!\equiv -x\!-\!1;\ \ \ \ \ 2(x^2\!-x\!-\!1) = \color{#b70}{(x\!+\!1)}\color{#0a0}{(2x\!-\!4)}+\color{#c00}2\,\to\\ 2[\![2]\!]\!-\!\color{#b70}{(x\!+\!1)}[\![3]\!] = [\![4]\!] &\color{#c00}{2\:\!f}&\!\!\!\equiv\, \color{#c00}{x^2\!+\!2x\!+\!3 = g_2\!-\!1}\\[-1em] \rm\small quotients\qquad\!\qquad &\rm\small remainders\ \ \ \ \end{array}$

Or since $\,p = x^3\!-\!5\,$ is prime we can use Gauss's inversion algorithm, so the last step is instead

$\!\!\begin{array}{rrl}[\![1]\!]\!-\!\color{#0aa}{(x\!+\!1)}[\![2]\!] = [\![3]\!] &\ \ \ \ \,\color{#0a0}{(2x\!-\!4)}f&\!\!\!\equiv -x\!-\!1; \ \ \ \ \ 2(x^3\!-\!5) = \smash[b]{\color{#b70}{\underbrace{(x^2\!+\!2x\!+\!4)}_{\large q(x) \,=\, g_2\!\!\!\!\!\!\!\!}}}\color{#0a0}{(2x\!-\!4)}+\color{#c00}6\,\to\\[.2em] 2[\![1]\!]\ -\ \color{#b70}{q(x)}\:\![\![3]\!] = [\![4]\!] &\color{#c00}{6f}&\!\!\!\equiv \color{#c00}{3(x^2\!+\!2x\!+\!3)\!=\!3(g_2\!-\!1)} \end{array}$

Gauss's Algorithm uses the remainder sequence $\,r_{n+1} = p\bmod r_n\,$ (vs. $\,r_{n+1} = r_{n-1}\bmod r_n\,$ in the Euclidean algorithm). Writing the above in equivalent fraction form, using $x^3\equiv 5,\,$ we get

$$f \equiv \dfrac{1}{x^2\!-\!x\!-\!1}\,\color{#0aa}{\dfrac{x\!+\!1}{x\!+\!1}}\equiv\dfrac{x\!+\!1}{4\!-\!2x} \equiv \dfrac{x\!+\!1}{4\!-\!2x}\:\!\color{#b70}{\dfrac{g_2}{g_2}}\equiv\color{#c00}{\dfrac{3(g_2\!-\!1)}{6}}\qquad $$

Our first method is exactly equivalent to that above - which is the same as lone student's answer after reducing lone's fractions $\!\bmod x^3\!-\!5$ (i.e. substitute $x^3\equiv 5)$ at every step (vs. at end). Note that the final step in lone's answer, i.e. scaling the fraction by $\,4\:\!\color{#b70}{q(x)}=\color{#08f}{(4^3\!-\!(2x)^3)/(4\!-\!2x)}\,$ is just another method of computing the final quotient $\:\!\color{#b70}{q(x)}$ when the divisor is linear $= x-a,\,$ and when $p = x^n - c.\,$ Namely, by the Polynomial Remainder Theorem we have

$$ p(x) = (x\!-\!a) q(x) + p(a) \,\Rightarrow\, \color{#b70}{q(x)} = \dfrac{p(x)-p(a)}{x-a} \left[= \color{#08f}{\dfrac{x^n-a^n}{x-a}}\ \ {\rm if}\ \ p = x^n-c\,\right]$$

Thus lone's method is a slight rearrangement of Gauss's inversion algorithm that works in the special case of inverting a quadratic $\,f\not\equiv 0\,$ modulo a prime $\ p = x^n-c.\,$ But Gauss's algorithm works more generally for any degree $f\not\equiv 0\,$ (and the extended Euclidean algorithm works more generally for coprime $p,f$ in any Euclidean domain, i.e. $p$ need not be prime - only coprime to $f)$

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  • $\begingroup$ Likely the simplest (general) way. $ $ OP is $\,\color{#c00}{{-}f \equiv -\frac{1}2(x^2\!+\!2x\!+\!3)},\ x= \sqrt[3]{5}\ $ (posted only for comparison to Galois-theoretic methods) $\endgroup$
    – Bill Dubuque
    Commented May 14, 2023 at 6:30
  • $\begingroup$ Equation Arithmetic: $\,\color{#0a0}{ a[\![j]\!]} + \color{#c00}{b[\![k]\!]} =: [\![\ell]\!]\,$ means that equation number $\,\ell\,$ is obtained by scaling equation $\color{#0a0}j$ by $\color{#0a0}a$ then adding it to the scaling of equation $\color{#c00}k$ by $\color{#c00}b.\,$ The sequence of coef's of $f$ are the remainders in the Euclidean algorithm for $\gcd(x^3\!-\!5,x^2\!-\!x\!-\!1),\,$ as explained here (for integers) $\endgroup$
    – Bill Dubuque
    Commented May 14, 2023 at 18:42
  • $\begingroup$ As e.g. here, we can use the following to rationalize any denom $\,x\!+\!y\!+\!z = a+b\sqrt[3]{5}+ c\sqrt[3]{5}^2$ $$x^3 + y^3 + z^3 - 3xyz \; = \; \left(x^2 + y^2 + z^2 - xy - xz - yz\right)\left(x+y+z\right)\qquad$$ $\endgroup$
    – Bill Dubuque
    Commented Jun 22 at 17:06
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Applying formulas for sum and difference of cubes, we can get immediately: $$\dfrac1{1+\sqrt[\large3]5-\sqrt[\large3]{25}} =\dfrac1{2-(1-\sqrt[\large3]5+\sqrt[\large3]{25})} =\dfrac{1+\sqrt[\large3]5}{2(1+\sqrt[\large3]5))-6} = \dfrac{\sqrt[\large3]5+1}{2(\sqrt[\large3]5-2)}$$ $$= \dfrac{(\sqrt[\large3]5+1)(\sqrt[\large3]{25}+2\sqrt[\large3]5+4)}{2(5-8)}=-\dfrac{9+6\sqrt[\large3]5+3\sqrt[\large3]{25}}6 =\color{green}{\mathbf{ -\dfrac12(\sqrt[\large3]{25}+2\sqrt[\large3]5+3)}.}$$

Evidently, is not a problem to express these calculations chain in the terms of the Galois's theory.

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    $\begingroup$ This is the same as lone student's answer (if you reduce $x^3\to 5$ at every step (vs. end), as I suggested yesterday in comments there). My answer shows it's a special case of Gauss's modular inversion algorithm. $\endgroup$
    – Bill Dubuque
    Commented May 16, 2023 at 17:07
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By polynomial long division, $$\frac{x^3-5}{-x^2+x+1}=-x-1+\frac{2x-4}{-x^2+x+1}=0\tag1$$ $$x\frac{x^3-5}{-x^2+x+1}=-x^2-x-2+\frac{-2x+2}{-x^2+x+1}=0\tag2$$ and by adding $(1)$ and $(2)$ we have $-x^2-2x-3-\frac{2}{-x^2+x+1}=0$ and $\frac{1}{-x^2+x+1}=-\frac12x^2-x-\frac32$.

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    $\begingroup$ This is just the extended Euclidean algorithm for $\,\gcd(x^3−5,f), f=−x^2+x+1,\,$ expressed in fraction form (usually done in continued fraction form). It is exactly the same as the first Extended Euclidean algorithm computation in my answer, if you combine your last two steps by multiplying equation $(1)$ by $\,\color{#c50}{x+1},\,$ instead of $\,x,\,$ i.e. $(2)+(1)=\color{#c50}x(1)+\color{#c50}1(1)=(x+1)(1),\,$ i.e. compute the quotient $−2(−x^2+x+1)\div\color{#0a0}{(2x−4)}=\color{#c50}{x+1}\,$ in one (vs. your two) steps, as in equation $[\![4]\!]$ in my answer. $\ \ $ $\endgroup$
    – Bill Dubuque
    Commented May 18, 2023 at 10:16
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    $\begingroup$ Note: in my answer I compute the inverse of the negation $−f,\,$ so signs differ. Also, your final result upon adding is not $=0$ but is $\equiv 0\pmod{\!x^3-5}\,$ or, equivalently, it is $=0$ when evaluated at $\,x = \sqrt[3]5.\ \ $ $\endgroup$
    – Bill Dubuque
    Commented May 18, 2023 at 10:22
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Let $x = \sqrt[3]{5}$ and let $y = \frac{1}{1 + x - x^2}$ (the value you want to solve for). Then:

$$(1 + x - x^2)y = 1 \tag{1}$$

Multiplying by $x$ (and simplifying $x^3 = 5$) gives:

$$(x + x^2 - 5)y = x \tag{2}$$

Multiplying by $x$ again gives:

$$(x^2 + 5 - 5x)y = x^2 \tag{3}$$

And multiplying by $x$ one more time just takes us back to equation (1), except scaled by a factor of 5.

Let's consider a weighted combination of the three equations we have, for some constants $A$, $B$, and $C$.

$$A(x^2 + 5 - 5x) + B(x + x^2 - 5) + C(1 + x - x^2) = \frac{Ax^2 + Bx + C}{y}$$ $$(A + B - C)x^2 + (-5A+B+C)x + 5A - 5B + C = \frac{Ax^2 + Bx + C}{y}$$

Try to pick constants that make the left side as simple as possible. Let's eliminate the $x^2$ term by setting $C = A + B$.

$$(-4A+2B)x + 6A - 4B = \frac{Ax^2 + Bx + A + B}{y}$$

Now eliminate the $x$ term by setting $B = 2A$.

$$-2A = \frac{Ax^2 + 2Ax + 3A}{y}$$

And now the $A$'s just cancel out.

$$-2 = \frac{x^2 + 2x + 3}{y}$$ $$y = \frac{-x^2 - 2x - 3}{2}$$ $$\boxed{y = \frac{-\sqrt[3]{25} - 2\sqrt[3]{5} - 3}{2}} \approx -4.671985$$

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    $\begingroup$ This is equivalent to inverting $\, \beta = 1+x-x^2\,$ by computing its norm as the determinant of the multiplication by $\beta$ map (e.g. see here) or using resultants, as I mentioned a few hours ago in comments on dxiv's answer. $\endgroup$
    – Bill Dubuque
    Commented May 19, 2023 at 2:36
  • $\begingroup$ Simpler: to compute the inverse $\,y\,$ we use elimination on your equations $(1),(2),(3)$ of form $\, g_i\, y = f_i\,$ to decrease the degree of the $g_i$ (= denominator of $\,y = f_i/g_i),\,$ yielding $$\begin{align} \color{#0a0}{(4)}\ :=\ (1)+(2) &\ \ \ \ \ \ \ \ (2x-4)\, y =\ \ x+1\\[.2em] \color{#90f}{(5)}\ :=\ (1)+(3)&\ \ \ \ \ (-4x+6)\,y = x^2+1\\[.2em] (6):=2\color{#0a0}{(4)}+\color{#90f}{(5)}&\qquad\quad\ \ \ \ \ \:\!\color{#c00}{{-}2\, y = x^2+2x+3}.\ \ \ \bf\small QED \end{align}$$ $\endgroup$
    – Bill Dubuque
    Commented May 19, 2023 at 19:13
  • 1
    $\begingroup$ This is very closely related to inversion by the extended Euclidean algorithm (cf. my answer) where we use Euclidean division (with remainder) to find smaller degree denominators. $\endgroup$
    – Bill Dubuque
    Commented May 19, 2023 at 19:13
0
$\begingroup$

Yet an other solution, that covers the general case: $$ \bbox[lightblue]{\qquad \xi=\xi(a) :=\frac 1{1+\sqrt[3]a-\sqrt[3]{a^2}} =\frac 1{1+A-A^2} \ , \qquad A:=\sqrt[3]a\ .\qquad } $$ The problem comes with $a=5$, the solution works in a more general case. But let us focus on this special $a=5$. Some motivation first. The number $\xi$ above lives in the field of algebraic numbers $\Bbb Q(\sqrt[3]{a}) =\Bbb Q(A)$. Which is not a Galois field. We may pass to a Galois closure $K$ of it, and then the norm $N_{K:\Bbb Q}(1+A-A^2)$ is:

  • a rational number of the one side,
  • a product of all conjugates of $\alpha_1:=1+\sqrt[3]{a}-\sqrt[3]{a^2} =1+A-A^2$ in $K$, which are $\alpha_1$, the same number, and some other numbers $\alpha_2,\alpha_3,\dots$

And from here, we already know what to do, we have $\xi=\frac 1{\alpha_1}=\frac {\phantom{\alpha_1}\alpha_2\alpha_3\dots}{\alpha_1\alpha_2\alpha_3\dots}=\frac{\alpha_2\alpha_3\dots}{N_{K:\Bbb Q}(\alpha_1)}$, and that norm in the denominator is a rational number.

Sometimes, it is enough to use only some of the conjugates, and here is so the case. We can start the computation / the solution:

Let $\varepsilon$ be a primitive root of order three of the unit, e.g. $\frac 12(-1+\sqrt{-3})$. We need only $\varepsilon^2+\varepsilon+1=0$. Consider the following numbers, obtained from $\alpha_1$ by (formally) replacing $\sqrt[3]{a}$ by itself, by $\varepsilon\sqrt[3]a$, and by $\varepsilon^2\sqrt[3]a$: $$ \begin{aligned} \alpha_1 &&&= 1+A-A^2 &&= 1+\sqrt[3]{a}-\sqrt[3]{a^2}\ ,\\ \alpha_2 &&&= 1+\varepsilon A-\varepsilon^2 A^2 && =1+\varepsilon\sqrt[3]{a}-\varepsilon^2\sqrt[3]{a^2}\ ,\\ \alpha_3 &&&= 1+\varepsilon^2 A-\varepsilon A^2 && =1+\varepsilon^2\sqrt[3]{a}-\varepsilon\sqrt[3]{a^2}\ . \end{aligned} $$ Recall the identity: $$ \begin{aligned} X^3 + Y^3 + Z^3 &-3XYZ \\ &=(X+Y+Z)(X^2 +Y^2+Z^2-YZ-ZX-XY) \\ &=(X+Y+Z)(X+\varepsilon Y+\varepsilon^2 Z)(X+\varepsilon^2 Y+\varepsilon Z) \ . \\[3mm] &\qquad\text{ Apply it for:}\\ X&=1\ ,\\ Y&=A=\sqrt[3]{a}\ ,\\ Z&=-A^2=-\sqrt[3]{a^2}\ .\\ &\qquad\text{ Then we get:}\\[3mm] \alpha_1\alpha_2\alpha_3 &=X^3+Y^3+Z^3-3XYZ=1+a-a^2+3a=1+4a-a^2\in\Bbb Q\ ,\\ \alpha_2\alpha_3 &=X^2+Y^2+Z^2-YZ-ZX-XY \\ &=1+A^2+A^4 +A^3+A^2-A\\ &=1+A^2+aA+a+A^2-A\\ &=(1+a)+(a-1)A+2A^2\ . \end{aligned} $$

For $\bbox[yellow]{\ a=5\ }$ we obtain: $$ \begin{aligned} \xi=\xi(5) &= \frac1{1+\sqrt[3]5-\sqrt[3]{25}} = \frac{\alpha_2\alpha_3}{\alpha_1\alpha_2\alpha_3} = \frac{6+4A+2A^2}{1+20-25} = -\frac 14(6+4A+2A^2) \\ &=-\frac 12(3+2A+A^2) =\bbox[yellow]{\ -\frac 12\left(\ 3 + 2\sqrt[3]5 + \sqrt[3]{25}\ \right)\ }\ . \end{aligned} $$

Computer check: Here using sage.

sage: R.<x> = PolynomialRing(QQ)
sage: K.<A> = NumberField(x^3 - 5)
sage: 1/(1 + A - A^2)
-1/2*A^2 - A - 3/2
$\endgroup$
1
  • $\begingroup$ Galois theoretical arguments have been already mentioned in one solution, and in many comments. Initially, as i saw this direction already mentioned, i was less attracted to double a solution. However, seeing the many other solutions - where the landscape becomes computationally really dirty, without being quick & dirty, i decided to post the above showing: $(1)$ what to do in the more general case, e.g. with $$\frac1{s+tA+uA^2}\ , \ A=\sqrt[3]a\ ,\ a,s,t,u\in\Bbb Z\ ,$$ and $(2)$ give the identity - which also solves the cubic, $X^3+Y^3+Z^3-3XYZ=\dots$ and is the core of the computation(s) $\endgroup$
    – dan_fulea
    Commented May 19, 2023 at 13:29

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