Hello,
Not it isn't.
$\displaystyle a^{{\color{red}n}*k}$ (I suppose it's n, not a) is equal to $\displaystyle (a^n)^k=(a^k)^n$
The only thing you can do is to factor by $\displaystyle a^k$, if $\displaystyle k<n$ :
$\displaystyle a^n+a^k=a^k \left(a^{n-k}+1 \right)$
No, you're not "going around in circles". Once you have found the two solutions $\displaystyle u_1,\,u_2$ of $\displaystyle u^2+u-6=0$, you can solve for $\displaystyle x$ : $\displaystyle \sqrt[4]{x^3+8}=u_1 \Rightarrow x^3+8=u_1^4$ and so on...
Be careful about $\displaystyle \Delta=b^2-4ac=1^2-4\times 1\times (-6)\neq 40 $
I'm referring to post #7: I don't understand where you've got the results from:
$\displaystyle u^2+u-6=0~\implies~u = -3~\vee~u=2$
And therefore
$\displaystyle \sqrt[4]{x^3+8} = -3$ that means there doesn't exist a real x to satisfy this equation
or
$\displaystyle \sqrt[4]{x^3+8} = 2~\implies~x^3+8=16~\implies~x^3=8~\implies~\boxed {x=2}$