If $x^3+\frac1{x^2}=1$, what is $x^3+\frac1{x^3}$?

Question

1. $x^3 + \frac1{x^2} = 1$. Then, $x^3 + \frac1{x^3} = ~?$

2. $p + \frac1{p^2} = 47$. Then, $p + \frac1p = ~?$

Answer 1

For the second problem (in the form: given that $p+\frac{1}{p^2}=47$, find $p+\frac1p$), we can show that in fact it's impossible for $p+\frac1p$ to be a rational number.

First, multiply the left-hand equation by $p^2$ to rewrite it as $f(p) = p^3-47p^2+1=0$. Now, $f$ is irreducible over $\mathbb{Q}$ (by the rational root theorem it has no rational roots, and since it's a cubic polynomial, if it were going to factor at least one of the roots would be linear). What's more, all three roots are real: $f(p)\to-\infty$ as $p\to-\infty$, $f(0)\gt 0$ by inspection, $f(46)\lt 0$ also by inspection (it's $46^2\cdot (46-47)+1$), and $f(p)\to\infty$ as $p\to\infty$ so we get roots in each of the intervals $(-\infty, 0)$, $(0,46)$ and $(46, \infty)$. This means that there's no 'clean' formula for any of the roots, just a Cardano-style formula.

Now, suppose that $p+\frac1p = \frac ab$, for some $a$ and $b$. Then this can be rewritten as $bp^2+b=ap$, or $g(p) = bp^2-ap+b=0$; but this implies that the cubic $f$ factors over $\mathbb{Q}$ as $f=gh$ for some linear $h$, which we've already shown impossible. Therefore, the expression $p+\frac1p$ can't possibly be rational.

(In fact, I believe this shows something more, namely that the expression can't even be a quadratic surd — because the extensions over $\mathbb{Q}$ have to be of different degrees — but my Galois skills aren't quite strong enough to be confident in that.)

Note that if that problem were instead to find $p+\frac1p$ given that $p^2+\frac1{p^2}=47$, then it has a much cleaner answer: set $x=p+\frac1p$, multiply out to find $x^2$ and substitute the known information about $p^2+\frac1{p^2}$ to get an easy equation for $x$. A similar approach will work for the first problem if we assume that the condition is in fact $x^2+\frac1{x^2}=1$ and the goal is to find $x^3+\frac1{x^3}$; in this case, we can set $x+\frac1x=c$. Then $c^2=x^2+2+\frac1{x^2}$ and $c^3=x^3+3x+3\frac1x+\frac1{x^3} = x^3+\frac1{x^3}+3c$. The first of these, along with the given information, will find you $c$, and you can then plug that information into the latter to find $x^3+\frac1{x^3}$. I won't spoil the surprise, but it's got a cute enough catch that I suspect it must have been the intended version of the problem!