Given that $\displaystyle x+\frac{1}{x}$ is a prime number.
Can $\displaystyle x^3+\frac{1}{x^3}$ also be a prime number?
Hello, Also sprach Zarathustra!
$\displaystyle \text{Given: }\,x+\dfrac{1}{x}\,\text{ is a prime number.}$
$\displaystyle \text{Can }\:x^3+\dfrac{1}{x^3}\,\text{ also be a prime number?}$
The only case is: .$\displaystyle x = 1.$
$\displaystyle \begin{array}{cccccccc}
\text{Suppose } x\ne1: & \qquad\qquad x + \dfrac{1}{x} &=& p \\ \\[-2mm]
\text{Cube both sides:} & \qquad\qquad \left(x + \dfrac{1}{x}\right)^3 &=& p^3 \\ \\[-2mm]
& x^3 + 3x + \dfrac{3}{x} + \dfrac{1}{x^3} &=& p^3 \\ \\[-2mm]
& \left(x^3 + \dfrac{1}{x^3}\right) + 3\underbrace{\left(x + \dfrac{1}{x}\right)}_{\text{This is }p} &=& p^3 \end{array}$
. . . . . . . . . . . . . . . . . . . .$\displaystyle \begin{array}{cccccc}
\left(x^3 + \dfrac{1}{x^3}\right) + 3p &=& p^3 \\ \\[-3mm]
\qquad x^3 + \dfrac{1}{x^3} &=& p^3 - 3p \\ \\[-3mm]
\qquad x^3 + \dfrac{1}{x^3} &=& \underbrace{p(p^2-3)}_{\text{composite}}
\end{array}$