1. ## Primes...

Given that $x+\frac{1}{x}$ is a prime number.

Can $x^3+\frac{1}{x^3}$ also be a prime number?

2. Originally Posted by Also sprach Zarathustra
Given that $x+\frac{1}{x}$ is a prime number.

Can $x^3+\frac{1}{x^3}$ also be a prime number?
Let $a=x^3+\frac{1}{x^3}$ and $p=x+\frac{1}{x}$.

We have $p^3 = a+3p$, hence $p | a$, hence no!

3. One point to Bruno J.

4. except the case $x = 1$

$p^2 - 3$ may equal $1$

5. Hello, Also sprach Zarathustra!

$\text{Given: }\,x+\dfrac{1}{x}\,\text{ is a prime number.}$

$\text{Can }\:x^3+\dfrac{1}{x^3}\,\text{ also be a prime number?}$

The only case is: . $x = 1.$

$\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}$

. . . . . . . . . . . . . . . . . . . . $\begin{array}{cccccc}
\left(x^3 + \dfrac{1}{x^3}\right) + 3p &=& p^3 \\ \\[-3mm]
\qquad x^3 + \dfrac{1}{x^3} &=& p^3 - 3p \\ \\[-3mm]
\qquad x^3 + \dfrac{1}{x^3} &=& \underbrace{p(p^2-3)}_{\text{composite}}
\end{array}$

6. Soroban you have already 2 points.

7. Originally Posted by Bruno J.
Let $a=x^3+\frac{1}{x^3}$ and $p=x+\frac{1}{x}$.

We have $p^3 = a+3p$, hence $p | a$, hence no!
... and also $a>p$.

8. Originally Posted by melese
... and also $a>p$.
Yeah I forgot that part, as others pointed out!