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.
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 $6$.
Either way, (1) is positive when $|t|>2$.
Thus, the only root of $f(t)=1$ outside of $(-2,2)$ is $t=2$.