Prove or disprove the following property: if $r$ is any non-zero rational number, then the real number $$x=(1+r)^{1/3}+(1-r)^{1/3}$$ is irrational.
Can I get some hints/guidance to solve this problem? Thanks in advance.
Attempt:
Initial idea was by assuming $x$ and $r$ are rational numbers with $r\neq 0$. Let $y_1=(1+r)^{1/3}$ and $y_2=(1-r)^{1/3}$, so we can get $$\frac{x^3-2}{3x}=y_1y_2$$
The equation for $x$ to check on is
$$x = \sqrt[3]{1 + r} + \sqrt[3]{1 - r} \tag{1}\label{eq1A}$$
By cubing both sides, along with that $\sqrt[3]{1 + r}(\sqrt[3]{1 - r}) = \sqrt[3]{1 - r^2}$ and Sai Mehta's comment suggestion, we get that
$$x^3 = (1 + r) + 3\sqrt[3]{1 + r}(\sqrt[3]{1 - r})(\sqrt[3]{1 + r} + \sqrt[3]{1 - r}) + (1 - r) = 3x\sqrt[3]{1 - r^2} + 2 \tag{2}\label{eq2A}$$
Using what you've shown as your initial idea, we can rearrange \eqref{eq2A} to get
$$\frac{x^3 - 2}{3x} = \sqrt[3]{1 - r^2} \tag{3}\label{eq3A}$$
Assuming $x$ is a rational number, then so is $\sqrt[3]{1 - r^2}$. Thus, if $r = \frac{a}{b}$ for integers $a \neq 0$, $b \gt 0$ and $\gcd(a, b) = 1$, then there exists integers $c$ and $d$ with $d \gt 0$ and $\gcd(c, d) = 1$ where
$$\sqrt[3]{1 - r^2} = \sqrt[3]{\frac{b^2 - a^2}{b^2}} = \frac{c}{d} \;\;\to\;\; \frac{b^2 - a^2}{b^2} = \frac{c^3}{d^3} \tag{4}\label{eq4A}$$
Due to the denominators being positive integers and the coprime conditions, the two numerators and two denominators are equal to each other, i.e.,
$$b^2 - a^2 = c^3, \;\;\; b^2 = d^3 \tag{5}\label{eq5A}$$
With $b^2 = d^3$, note that for any prime factor $p$ of $b$ and $d$, if $j$ is the exponent of it in $b$ and $k$ is that in $d$, we have $2j = 3k$. Thus, $3 \mid j$ and $2 \mid k$, i.e., $6 \mid (2j = 3k)$. Since this applies to all prime factors $p$ of $b$ and $d$, this means there's an integer $e$ where $b^2 = d^3 = e^6$ (note that, in general, if $b^m = d^n$, there's an integer $f$ where $b^m = d^n = f^{\operatorname{lcm}(m,n)}$). Thus, the first part of \eqref{eq5A} becomes
$$e^6 - a^2 = c^3 \;\;\to\;\; a^2 + c^3 = e^6 \tag{6}\label{eq6A}$$
Since $r \neq 0$, then $a \neq 0$. Also, $c = 0$ means that $r = \pm 1$, but then $x = \sqrt[3]{2}$ which is not rational. Thus, $a$, $c$ and $e$ are non-zero integers. The answer to No positive integer solution for $x^2 + y^3 = z^6$?, using the solution in the AoPS thread Diophantine equation of high degree, shows that for $x^2 + y^3 = z^3$
The only solutions of this equation in nonzero integers are $(x,y,z)=(\pm 3k^3,-2k^2,\pm k)$, where $k$ is nonzero.
The coprime requirement means that $k = \pm 1$, so $a^2 = (\pm 3(\pm 1)^3)^2 = 9$ and $b^2 = 1$. Thus, from \eqref{eq3A}, we get
$$\frac{x^3 - 2}{3x} = \sqrt[3]{1 - 9} = -2 \;\;\to\;\; x^3 + 6x - 2 = 0 \tag{7}\label{eq7A}$$
Since this is a cubic polynomial with integer coefficients, then the rational root theorem states the only possible rational roots are $x = \pm 1$ or $x = \pm 2$. Since $x$ must be even, we can just check that the LHS for $x = 2$ is $18$ and for $x = -2$ it's $-22$.
Thus, $x$ can't be rational. This shows that, in \eqref{eq1A}, $x$ is irrational for all rational $r \neq 0$.
