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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.

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By inspection, $x=1$ is a solution. –  Per Manne Oct 19 '12 at 15:27
That should be $3(t^4-4t^2+3)^3$ on the right. (Not that I've checked the math that got $t^4-4t^2+3$, but clearly the exponent is wrong.) –  Thomas Andrews Oct 19 '12 at 15:45
Not sure that you can do this analytically, seeing the results: tinyurl.com/95l2p8f –  Eric Angle Oct 19 '12 at 15:47
@EricAngle, given that solution, you might be able to prove analytically there is only one real solution. Given that the real solution is $x=1$ –  Thomas Andrews Oct 19 '12 at 15:50
Alpha finds 12 roots for the equation in $t$, with only $t=2$ corresponding to a real solution for $x$. The other three real $t$ solutions are less than $2$ in absolute value. –  Ross Millikan Oct 19 '12 at 15:51
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1 Answer

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 $t=2$.

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