How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

Question

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.

Answer 1

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 $g^{-1}\,$: $$ \begin{align} &&4g^{-1} &\equiv -g^2 + 3 g - 18 \\ &&&\equiv -(1+x-x^2)^2 + 3 (1+x-x^2) - 18 \\ &&&\equiv -x^4 + 2 x^3 - 2 x^2 + x - 16 \\ &&&\equiv -(x^3 - 5)(x - 2) -2 (x^2 + 2 x + 3) \\ &&&\equiv -2 (x^2 + 2 x + 3) \pmod{x^3 - 5} \\ \implies\;\; &&(1+x-x^2)^{-1}\;\; &\equiv - \frac{1}{2}(3 + 2x + x^2) \pmod{x^3 - 5} \end{align} $$

Answer 2

My goal will be to simply derive expressions in the denominator in terms of $\thinspace x^{3n}\thinspace . $

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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 :

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

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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 .

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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 $\thinspace x^{3n}$ :

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

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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 :

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

Answer 3

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 $$ \dfrac{1}{1+\sqrt[3]{5}-\sqrt[3]{25}}=\dfrac{1}{-z^2+z+1}=-\dfrac{1}{2}z^2-z-\dfrac{3}{2}=-\dfrac{1}{2}\sqrt[3]{25}-\sqrt[3]{5}-\dfrac{3}{2} $$

Answer 4

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

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

Answer 5

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

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

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).

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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}\ $ i.e. in equation form:

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

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

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

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

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

The method in lone student's answer is exactly equivalent to that above - as can be seen by 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 $f)$

Answer 6

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.

Answer 7

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 $\frac{1}{-x^2+x+1}=-\frac12x^2-x-\frac32$.

Answer 8

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.

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

Answer 9

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:

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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: $$ \begin{aligned} X^3 + Y^3 + Z^3 &-3XYZ \\ &=(X+Y+Z)(X^2 +Y^2+Z^2-YZ-ZX-XY) \\ &=(X+Y+Z)(X+\varepsilon Y+\varepsilon^2 Z)(X+\varepsilon^2 Y+\varepsilon Z) \ . \\[3mm] &\qquad\text{ Apply it for:}\\ X&=1\ ,\\ Y&=A=\sqrt[3]{a}\ ,\\ Z&=-A^2=-\sqrt[3]{a^2}\ .\\ &\qquad\text{ Then we get:}\\[3mm] \alpha_1\alpha_2\alpha_3 &=X^3+Y^3+Z^3-3XYZ=1+a-a^2+3a=1+4a-a^2\in\Bbb Q\ ,\\ \alpha_2\alpha_3 &=X^2+Y^2+Z^2-YZ-ZX-XY \\ &=1+A^2+A^4 +A^3+A^2-A\\ &=1+A^2+aA+a+A^2-A\\ &=(1+a)+(a-1)A+2A^2\ . \end{aligned} $$

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

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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