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).

  • 3
    $\begingroup$ Since the equation is effectively a cubic, there is an exact form...but it certainly isn't pretty. See this $\endgroup$
    – lulu
    Feb 21 '20 at 12:42
  • $\begingroup$ Thank you for your answer, sadly it won't let me check on the maths behind it. $\endgroup$ Feb 21 '20 at 12:49
  • 1
    $\begingroup$ You probably need to look up Cardano's solution of the cubic, which is no doubt what WA is using. $\endgroup$ Feb 21 '20 at 12:50
  • $\begingroup$ 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). $\endgroup$
    – lulu
    Feb 21 '20 at 12:55
  • $\begingroup$ Are you asking for a derivation of Cardano's formula which is the formula that gives you the real root of the cubic? $\endgroup$
    – bjorn93
    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

$$x = \frac{\sqrt[4]{3}}{\sqrt2}\left[ \sinh\left(\frac13\sinh^{-1}\frac{3\sqrt3}2\right)\right]^{-\frac12}=1.21$$


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


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