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.

Let $a_k=x^k+\frac1{x^k}$， we have
$\begin{cases}a_1a_4=a_5+a_3\\a_2a_3=a_5+a_1\\a_1a_3=a_4+a_2\end{cases}$
By writing $a_3=u,a_4=v$, we could rewrite the fomula as
$\begin{cases}va_1+0a_2-a_5=u\\a_1-ua_2+a_5=0\\ua_1-a_2+0a_5=v\end{cases}$
Solving the linear equation of $a_1,a_2,a_5$ we could get they're all rational number
