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I need to rationalize $\displaystyle\frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2}$

I'm given what I need to rationalize it, namely $\displaystyle\frac{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}$

But I fail to see the motivation behind this. Can someone give a more general explanation of how to come to this expression?

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4 Answers 4

up vote 5 down vote accepted

You can multiply it by $1+d\sqrt[3]2+e\sqrt[3]4$ and insist the product not have any factors of $\sqrt[3]2$ or $\sqrt[3]4$. So $$(a+b\sqrt[3]2+c\sqrt[3]4)(1+d\sqrt[3]2+e\sqrt[3]4)=\\a+ebe+2cd+\sqrt[3]2(ad+b+2ce)+\sqrt[3]4(ae+bd+c)$$

So we need $ad+b+2ce=0, ae+bd+c=0$, giving $$e=-\frac {ad+b}{2c}\\-a^2d-ab+2cbd+2c^2=0\\(a^2-2bc)d=2c^2-ab\\(a^2-2bc)e=b^2-ac$$ Then we make the leading term $a^2-bc$ to clear the fractions

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Heh. Beat me to the punch. +1 –  Rick Decker Feb 5 '13 at 1:45

You can also rationalize this by making use of the algebraic identity

$$x^3 + y^3 + z^3 - 3xyz \; = \; \left(x^2 + y^2 + z^2 - xy - xz - yz\right)\left(x+y+z\right)$$

You can view this as analogous to binomial square root rationalization, in which having a denominator of the form $x+y,$ where $x^2$ and $y^2$ are free of radicals, can be freed of radicals by multiplying both numerator and denominator by $x-y.$ In your situation, $x+y+z$ is what you have in the denominator and what you do is multiply both numerator and denominator by the "conjugate" $x^2 + y^2 + z^2 - xy - xz - yz.$ This identity doesn't take care of rationalizing a general trinomial made up of cube roots, by the way. For this there is a much more complicated identity that will work. See this 16 November 2010 AP-calculus post at Math Forum for a discussion of how to obtain such an identity.

Incidentally, this algebraic identity has a lot of uses, the best known probably being one of the ways of obtaining the cubic formula for solving cubic equations. This identity makes a lot of appearances in 19th century algebra texts and more about it can be found in the following sci.math thread I started on 23 April 2009: Factorization of a^3 + b^3 + c^3 - 3abc. I've assembled over a hundred old references to it since then, and one day I might write a historical survey of its mathematical and educational appearances in 19th century literature. It's also a standard identity for those serious about mathematical olympiad problems.

A few months after those sci.math posts I discovered the following short survey paper on this algebraic identity:

Desmond MacHale, My favourite polynomial, The Mathematical Gazette 75 #472 (June 1991), 157-165.

See also Mark B. Villarino's recent (2 January 2013) paper A cubic surface of revolution.

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You rationalize by using Euclid's algorithm as used to prove Bézout's identity.

Let $\alpha = \sqrt[3]{2}$. We have obviously to assume $a+b\sqrt[3]2 + c(\sqrt[3]2)^2 = a+b\alpha + c \alpha^2 \ne 0$.

Consider the polynomial $0 \ne a + b x + c x^2 \in \mathbb{Q}[x]$. Since $x^3 - 2$ is irreducible in $\mathbb{Q}[x]$, for the gcd we have $(a + b x + c x^2, x^3 - 2) = 1$. Use Euclid's algorithm to find polynomials $u, v \in \mathbb{Q}[x]$ such that $(a + b x + c x^2) \cdot u + (x^3 - 2) \cdot v = 1$. Now evaluate this for $x = \alpha$ to obtain $(a + b \alpha+ c\alpha^2) \cdot u(\alpha) = 1$, so that \begin{equation} \frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2} = \frac{1}{a + b \alpha+ c \alpha^2} = u(\alpha). \end{equation}

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I prefer this approach. Magic Bezout's identity. +1 –  Babak S. Feb 6 '13 at 2:44

You rationalize by multiplying by the conjugates. The conjugates of $\root3\of2$ are $\rho_1\root3\of2$ and $\rho_2\root3\of2$ where $\rho_1=e^{2\pi i/3}$ and $\rho_2=\rho_1^2=e^{4\pi i/3}$. So, you want to multiply top and bottom by $$(a+b\rho_1\root3\of2+c\rho_2\root3\of2^2)(a+b\rho_2\root3\of2+c\rho_1\root3\of2^2)$$ Multiply all that out, and simplify by using $\rho_1\rho_2=1$ and $\rho_2^2=\rho_1$ and maybe $1+\rho_1+\rho_2=0$, and with any luck you'll get what you're looking for.

This works in general, in that to rationalize you can always multiply by all the conjugates. The bad news is that it's not always so easy to find simple expressions for the conjugates.

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