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Can someone please tell me how to write $\frac{1}{7+\sqrt[3]{2}}$ as a linear combination of $1$, $\sqrt[3]{2}$ and $\sqrt[3]{2}^2$using only coefficients in $\mathbb{Q}$ ?

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Let $\omega = \sqrt[3]{2}$, it has minimal polynomial $\omega^3 - 2$.

Like when you divide complex numbers ($\frac{1}{a + ib} = \frac{a - ib}{a^2 + b^2}$) an important part of this is the norm (recall $N(a + ib) = a^2 + b^2$) but unlike the complex case, the automorphisms of the field don't help (because there are not automorphisms of $\mathbb Q(\omega)$!).

The definition of the norm in the general (non-Galois) case is not by multiplying all the conjugates together but the determinant of the linear operator "multiplication by $a + \omega b + \omega^2 c$". So write [a,b,c] as the vector representation of that number. Multiplication by it is represented by the matrix [a,b,c; 2c,a,b; 2b,2c,a] (not that [1,0,0]*that = [a,b,c]) or $$\left( \begin{array}{ccc} a & b & c \\ 2c & a & b \\ 2b & 2c & a \end{array} \right)$$ so we have $N(a + \omega b + \omega^2 c) = \text{det}(\text{that matrix}) = a^3 - 6abc + 2b^3 + 4c^3.$ So dividing by this number is represented by the inverse of that matrix, and we know $\text{det}(M)M^{-1}$ has a nice simple form so let's compute [1,0,0]*det(m)/m = [a^2 - 2*c*b, -b*a + 2*c^2, -c*a + b^2] to find $N(a + \omega b + \omega^2 c)/(a + \omega b + \omega^2 c) = (a^2-2bc) + (-ab + 2c)\omega + (-ac + b^2)\omega^2.$

Putting $a = 7$, $b = 1$, $c = 0$ this gives $$\frac{1}{7 + \omega} = \frac{49 - 7 \omega + \omega^2}{345}.$$

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  • $\begingroup$ Wow, great answer! (although some parts yet elude me; but I will get back to them) $\endgroup$ – user47574 Oct 30 '12 at 19:21
  • $\begingroup$ @user47574, I'm happy to explain parts more if you like. By the way this works for finding 1/x in any algebraic extension. $\endgroup$ – sperners lemma Oct 30 '12 at 19:29
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$$\dfrac1{a+b} = \dfrac{a^2 -ab +b^2}{a^3 + b^3}$$ Take $a = 7$ and $b=\sqrt[3]{2}$ to get what you want.

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  • $\begingroup$ Actually I meant $\mathbb{Q}(\sqrt[3]{2})$, sorry. $\endgroup$ – user47574 Oct 30 '12 at 18:38
  • $\begingroup$ That is magic, why did the factoring of a^3 + b^3 work? Are there other times when you can do this? $\endgroup$ – sperners lemma Oct 30 '12 at 19:14
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    $\begingroup$ @sperner, it works because of the known factorization $A^3+B^3=(A+B)(A^2-AB+B^2)$. Your method works in general for a quantity of type $\alpha+\beta m^{1/3}+\gamma m^{2/3}$, but the binomiality of the denominator allows the simplification. $\endgroup$ – Lubin Oct 30 '12 at 19:32

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