Resolving $x^6-x^4-1=0$ (only interested in $x\simeq1.21$) I'm struggling to find the solution to $x^6-x^4-1=0$, so far I tried changing variable ($u=x^2$) to work around $u^3-u^2-1=0$ instead but I'm still having a hard time.
And again, I'm only interested in finding the exact expression of the value $x\simeq1.21$.
Also I'm sorry if I posted to the wrong site (there was this one or MathOverflow).
 A: Here is how to derive the exact solution for $x$. Let $u=\frac1{x^2}$. Then, the equation becomes
$$u^3+u-1=0\tag 1$$
Compare with the identity 
$$4\sinh^3a+3\sinh a -\sinh 3a =0$$
and recognize that 
$$\sinh a =  \frac{\sqrt3}2u,\>\>\>\>\>\sinh 3a = \frac{3\sqrt3}2$$
satisfy (1). Then, $a = \frac13\sinh^{-1}\frac{3\sqrt3}2$ and the solution for $u$ is
$$u = \frac2{\sqrt3}\sinh\left(\frac13\sinh^{-1}\frac{3\sqrt3}2\right)$$
Thus, the solution for $x$ is
$$x = \frac{\sqrt[4]{3}}{\sqrt2}\left[ \sinh\left(\frac13\sinh^{-1}\frac{3\sqrt3}2\right)\right]^{-\frac12}=1.21$$
A: COMMENT.- Pay attention to Lulú's comment above. Numerical methods give faster and sufficiently approximate results for you. Here I show you is a sample (maybe interesting for you) of how the root you want determines all the other five roots.
Since $a$ is root if and only if $-a$ is root, the polynomial  $x ^ 6-x ^ 4-1$ is divisible by $x ^ 2-a ^ 2$ and then you have
$$x^6-x^4-1=(x^2-a^2)(x^4+(a^2-1)x^2+a^2(a^2-1))$$
Then $$x=\pm\sqrt{\frac{1-a^2\pm\sqrt{-3a^4+2a^2+1}}{2}}$$ gives four solutions and $x=\pm\sqrt{a^2}$ are the other two ones.
A: Let $t=x^2$ to face
$$t^3-t^2-1=0$$
Solving the cubic for $t$, there is only one real root. Using the hyperbolic method, you have
$$t=\frac{1}{3}\left(1+2 \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\frac{29}{2}\right)\right)\right)\approx 1.4655712$$
