<s>[INST] Let $x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$ are rational numbers. Show that $x+\frac{1}{x}$ is rational. 
$x^3+\dfrac{1}{x^3}$ and $x^4+\dfrac{1}{x^4}$ are rational numbers. Prove that $x+\dfrac{1}{x}$ is rational number.

My solution:

$x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right)$
$x^4+\dfrac{1}{x^4}=x^4+2+\dfrac{1}{x^4}-2=\left(x^2+\dfrac{1}{x^2}\right)^2-1-1=\left(x^2+\dfrac{1}{x^2}-1\right)\left(x^2+\dfrac{1}{x^2}+1\right)-1$
Because $x^4+\dfrac{1}{x^4}$ is rational number so
$\left(x^2+\dfrac{1}{x^2}-1\right)$ is rational number too and
$x^3+\dfrac{1}{x^3}$ is rational number so
$\left(x+\dfrac{1}{x}\right)$ is rational number.

Am I wrong? Please check my solution, thank you.
 [/INST] I constructed a possible elementary algebraic way:
Let,
$$\begin{cases}x+\frac 1x=v,\thinspace v \not \in \mathbb Q\\
x^2+\frac{1}{x^2}=u,\thinspace u\in\mathbb R\\
x^3+\frac{1}{x^3}=p,\thinspace p\in\mathbb Q\\
x^4+\frac{1}{x^4}=q,\thinspace q\in\mathbb Q\end{cases}$$
This implies,
$$pv=u+q\not\in \mathbb Q\implies u\not\in\mathbb Q\\$$
Then we have,
$$\begin{align}&uq=p^2-2+u\\
\implies &uq-u=p^2-2\\
\implies &u=\frac{p^2-2}{q-1}\in\mathbb Q\\
&\thinspace \text{A contradiction.}\end{align}$$
</s>