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If $\beta$ is the root of the equation $x^3-x-1=0$, find the value of $$(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}.$$

This is what I tried:

$x=\beta$ is a root of $x^3-x-1=0,$

so getting $\displaystyle \beta^3-\beta-1=0\Rightarrow \beta^2=\frac{\beta+1}{\beta}.$

Now $$3\beta^2-4\beta = 3\bigg(\frac{\beta+1}{\beta}\bigg)-4\beta.$$

Don't know how to continue.

Blue
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jacky
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2 Answers2

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The trick is that if we reduce $(x + a)^3$ modulo $x^3 - x - 1$, we'll get $3 a x^2 + (3 a^2 + 1) x + (a^3 + 1)$. Therefore, if $\beta$ is a root of $x^3 - x - 1$, $$(1 + \beta)^3 = 3 \beta^2 + 4 \beta + 2, \\ (1 - \beta)^3 = 3 \beta^2 - 4 \beta.$$ Then we need to define the cube root in such a way that $(z^3)^{1/3} = z$ for $z = 1 \pm \beta$.

Maxim
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  • Excellent! I should have tried checking whether $3\beta^2-4\beta$ and $3\beta^2+4\beta+2$ were cubes! – lhf Aug 21 '19 at 11:23
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I assume that $\beta$ is the real root of $x^3-x-1=0$ so that the cube roots are well defined.

Let $\mu = 3\beta^2-4\beta$ and $\nu = 3\beta^2+4\beta+2$. We seek $\tau = \mu^{\frac{1}{3}}+\nu^{\frac{1}{3}}$. As in this question, we have $$ \tau^3 = \mu+\nu+3(\mu\nu)^{\frac{1}{3}}\tau $$ Hoping that $\mu\nu = \beta-\beta^2$ is a cube in $\mathbb Q(\beta)$, we find after some work that $\beta-\beta^2=(1-\beta^2)^3$. Therefore, $$ \tau^3 = \mu+\nu+3(\mu\nu)^{\frac{1}{3}} = 6\beta^2+2+3(1-\beta^2)\tau $$ Now $\tau=2$ is a root. The other roots are complex because the discriminant of the quotient quadratic is $-12\beta^2$.

Therefore, $\tau=2$ is the only real root and the answer is $2$.

lhf
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    This answer used lots of hindsight and help from WA. I'd love to see a direct answer. – lhf Aug 20 '19 at 16:02
  • $\beta$ is inside $(0, 4/3)$. It's not possible to choose a branch of $z^{1/3}$ to make $\mu^{1/3} + \nu^{1/3}$ real. – Maxim Aug 20 '19 at 18:43
  • @Maxim I'm sorry, I don´t understand why that is the case. Could you expand on why that sum can´t be real please? – Fede Poncio Aug 20 '19 at 19:13
  • @FedePoncio Sorry, my claim wasn't quite correct. Since $\mu < 0$ and $\nu > 0$, we have to define $z^{1/3}$ to be negative for negative $z$ and positive for positive $z$. Then $\tau = 2$. This is a valid choice of $z^{1/3}$, just with a fancy branch cut structure. If we view $\mu^{1/3} + \nu^{1/3}$ as a multi-valued function, one of its values is $2$, it doesn't matter which $\beta$ we choose. But if we want to treat it as single-valued, we have to specify which cube roots we take for which $\beta$. – Maxim Aug 20 '19 at 19:47
  • @Maxim, my answer starts with the assumption that $\beta$ is real. The OP says "the root". – lhf Aug 21 '19 at 09:45
  • Sure, I was just confused about what "well defined" meant. For me, something like "assume that $\beta$ is the real root and that $\mu$ and $\nu$ are also real" would have been clearer. It doesn't matter how we choose $\beta$, the point is how we choose the cube roots. – Maxim Aug 21 '19 at 10:37