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If $x^3-\frac1{x^3}=108+76\sqrt2$, find $x-\frac1x$

LHS = $(x-\frac1x)(x^2+\frac1{x^2}+1)=(x-\frac1x)((x-\frac1x)^2+3)$

Now, maybe RHS needs to be factorized so that some comparisons can be made, but not able to do so.

Or maybe LHS can be written as $(x-\frac1x)^3+3(x-\frac1x)$. Now, RHS can be broken down into two terms. One could be the cube of one third of the other term, but not able to do this either.

Any ideas how to approach such questions?

aarbee
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  • Isn't this just a cubic equation in $y = x - \frac{1}{x}$ that can be solved by standard methods? Looks like you already did the trick to express the LHS in terms of what you want? – John Jun 05 '21 at 19:40
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    Try plugging in $(x-1/x)=a+b\sqrt{2}$? – Shengkai Li Jun 05 '21 at 19:41
  • Related: https://math.stackexchange.com/questions/987223/f-leftx-frac1x-right-x3x-3-find-fx ; more generally,see https://math.stackexchange.com/questions/678650/if-x-frac1x-5-find-x5-frac1x5?rq=1 –  Jun 05 '21 at 19:52

3 Answers3

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let $z = x - x^{-1}$; then we seek a "nice" solution of the form $z = a + b \sqrt{2}$ such that $z^3 + 3z = 108 + 76 \sqrt{2}$. This in turn implies $$a(3 + a^2 + 6b^2) = 108, \\ b(3 + 3a^2 + 2b^2) = 76.$$ If $a, b$ are integers, then we must have $a \in \{1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108\}$ and $b \in \{1, 2, 4, 19, 38, 76\}$ but clearly we can eliminate most of these possibilities, for if $a > 108^{1/3} > 4$ or $b > 38^{1/3} > 3$ neither equation can be met. This leaves easy casework to check, and we find that $a = 3, b = 2$ works, hence $z = 3 + 2\sqrt{2}$.

heropup
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  • But there are 2 solutions. This answer gives only 1 solution. – NN2 Jun 05 '21 at 19:52
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    @NN2 No. The solution of the cubic $z^3 + 3z = 108 + 76 \sqrt{2}$ has exactly one real solution, and two complex solutions, thus there is only one value for $x - 1/x$ that is real-valued. – heropup Jun 05 '21 at 19:56
  • Ok, I see now. thanks – NN2 Jun 05 '21 at 19:57
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Let $c = 108 + 76\sqrt{2}$ and $y = x^3 - 1/x^3$. Then $y(y^2+ 3) = y^3 + 3y = c$. This is a cubic equation whose real solution is given by

$$ y=\frac{\sqrt[3]{\sqrt{c^{2}+4}+c}}{\sqrt[3]{2}}-\frac{\sqrt[3]{2}}{\sqrt[3]{\sqrt{c^{2}+4}+c}} $$

Matheus Andrade
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Not an answer since it's not obvious how you would find this without already knowing the answer, but once you know the answer you can verify it easily by factoring correctly: $$ \left(\color{green}{x-\frac{1}{x}}\right)\left[\left(\color{green}{x-\frac{1}{x}}\right)^2+3\right] = x^3-\frac{1}{x^3}=108+76\sqrt{2} = \left(\color{green}{3 +2 \sqrt{2}}\right)\underbrace{\left[\left(\color{green}{3 +2 \sqrt{2}}\right)^2+3\right]}_{\color{blue}{12 \sqrt{2} +20}} $$

Robert Lee
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