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$$x=\sqrt[3]{p}+\sqrt[3]{q}$$

I'm trying to figure out some way to "rationalize" the previous equation, meaning to rewrite it purely in terms of whole number powers of $p$, $q$, and $x$. It seems quite simple but I've been stuck trying to do it. I'd appreciate anyone's help with this.

quote
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  • Not easy. If $p$ and $q$ are coprime, the polynomial has dgree $9$. – ajotatxe May 19 '15 at 17:45
  • Closely related: http://math.stackexchange.com/questions/1010973/degree-of-an-algebraic-number-over-a-field ; http://math.stackexchange.com/questions/359054 – MJD May 19 '15 at 18:14

3 Answers3

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Hint: Use the high school identity: $$(a+b)(a^2-ab+b^2)=a^3+b^3.$$

Bernard
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  • But that would leave $(\sqrt[3]{p^2}+\sqrt[3]{pq}+\sqrt[3]{q^2})x$ on the left hand side, wouldn't it? – quote May 19 '15 at 17:57
  • No: you write $a+b=\dots$. So really, what you rationalise is the numerator. Btw, it's $-\sqrt[3]{pq}$ that you should write. – Bernard May 19 '15 at 18:00
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Following up on Bernard's suggestion: $$x= \sqrt[3]{p}+\sqrt[3]{q}$$ $$x(\sqrt[3]{p^2}-\sqrt[3]{pq}+\sqrt[3]{q^2})= p+q$$ $$x((\sqrt[3]{p}+\sqrt[3]{q})^2-3\sqrt[3]{pq})= p+q$$ $$x(x^2-3\sqrt[3]{pq})= p+q$$ $$x^3-3x\sqrt[3]{pq}= p+q$$ $$3x\sqrt[3]{pq}= x^3-p-q$$ $$27x^3pq= (x^3-p-q)^3$$

user53970
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2

Another way (maybe it's the same as Bernard's, but I don't see how):

$${x\over \sqrt[3]{q}}=1+\sqrt[3]{p\over q}\\$$ $${x^3\over q}=1+{p\over q}+3\sqrt[3]{p\over q}\left(1+\sqrt[3]{p\over q}\right)=1+{p\over q}+3x\sqrt[3]{p\over q^2}$$

And you can continue from here.