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Can any one give me example of: rational functions $f, g$ and $h$ with rational coefficients such that

$$(f(x))^{3} + (g(x))^{3} + (h(x))^{3}=x$$

Also, if anyone knows a procedure for constructing such examples, I would be happy to learn them as well.

Added: Can we find rational functions $f$ and $g$ such that $(f(x))^3+(g(x))^3=x$

  • 1
    I don't know the answer, but by clearing denominators, you reduce this to $f(x)^3+g(x)^3+h(x)^3=x q(x)^3$, for *polynomials* with *integer coefficients*. This may be easier to think about. Setting $x=1$, you get an integer solution of $a^3+b^3+c^3=d^3$. This [paper](http://www.math.psu.edu/vstein/preprints/Rachel11.ps) is relevant then. Also, by setting $x=0$, you get Fermat's equation for exponent 3, which only has trivial solutions. In particular, $x$ divides one of $f$, $g$, $h$. Not that it helps...2011-05-05
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    Another observation is that (using lhf's notation) we must have all coefficients of $q(x)$ divisible by $3$. This is obtained by modulo 3 on both side, so we get $(f+g+h)^3=xq^3$ in the polynomial ring $\mathbb{F}_3[x]$, but $x$ is not a perfect cube, so this happens only when both sides are zero.2011-05-05
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    Start with $f(x)^3+g(x)^3+h(x)^3=x q(x)^3$. Suppose $q(x)$ has degree $k$. Then the RHS is asymptotic to $cx^{3k+1}$ as $x \to \infty$. But the LHS is asymptotic to $dx^{3m}$ unless there is cancellation of the highest term... By exponent 3 of Fermat, that means two of the terms have higher degree than the third, and their highest terms are negatives.2011-05-05

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