Solve for $x$: $\big(x^3+\frac{1}{x^3}+1\big)^4=3\big(x^4+\frac{1}{x^4}+1\big)^3$ Solve for $x$
$$\big(x^3+\frac{1}{x^3}+1\big)^4=3\big(x^4+\frac{1}{x^4}+1\big)^3$$
let $x+\frac{1}{x}=t$ the equation equivalent to $(t^3-3t+1)^4=3(t^4-4t^2+3)^3$
but it's very complicated. Thanks.

Let $u=t^4-4t^2+3$ and $v=t^3-3t+1$. To finish off the problem (over real numbers), it suffices to prove that the equation $3u^3=v^4$ has no solutions with $|t|\ge 2$ other than $t=2$. Differentiate the ratio $f(t)=3u^3 v^{-4}$: 
$$ f'(t)=\frac{d}{dt}(3u^3v^{-4})= 3u^2v^{-5}(3u'v-4uv') \tag1$$
The sign of $v$ is the same as the sign of $t$ when $|t|\ge 2$. It remains to find the sign of $3u'v-4uv'$. Direct computation shows 
$$3u'v-4uv' = 12(t^3-t^2-2t+3) \tag2$$ 
which is not as bad as one might have expected from subtracting two polynomials of degree $6$. 


*

*When $t \ge 2$ we have $t^3\ge 2t^2$, hence (2) is positive. 

*When $t\le -2$ we have $t^3\le -2t^2$, hence (2) is negative.


Either way, (1) is positive when $|t|>2$. 


*

*Since $f(2)=1$, it follows that $f(t)>1$ for $ t >2$.

*Since $\lim_{t\to-\infty}f(t)=3$,   it follows that $f(t)>3$ for $ t <-2$.


Thus, the only root of $f(t)=1$ outside of $(-2,2)$ is $t=2$.  
