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).

• Since the equation is effectively a cubic, there is an exact form...but it certainly isn't pretty. See this
– lulu
Feb 21 '20 at 12:42
• Thank you for your answer, sadly it won't let me check on the maths behind it. Feb 21 '20 at 12:49
• You probably need to look up Cardano's solution of the cubic, which is no doubt what WA is using. Feb 21 '20 at 12:50
• here is a discussion of the closed form for a cubic. It is straight forward, but quite messy. In practice, numerical methods tend to be preferred (though of course it depends on what you are looking for).
– lulu
Feb 21 '20 at 12:55
• Are you asking for a derivation of Cardano's formula which is the formula that gives you the real root of the cubic? Feb 21 '20 at 14:05

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

$$x = \frac{\sqrt{3}}{\sqrt2}\left[ \sinh\left(\frac13\sinh^{-1}\frac{3\sqrt3}2\right)\right]^{-\frac12}=1.21$$

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.

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 $$t=\frac{1}{3}\left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{29}{2}\right)\right)\right)\approx 1.4655712$$