Example of a rational function such that : $(f(x))^{3} + (g(x))^{3} + (h(x))^{3}=x$ 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 $(f(x))^3+(g(x))^3=x$

 A: $(a^4-2a)^3+(a^3+1)^3+(2a^3-1)^3=(a^4+a)^3$
There are more at http://sites.google.com/site/tpiezas/010 Section 2. 
This is Number Theory, not real analysis. 
A: You may be interested to know identities are known that can bring this higher:
$a_1(x)^3+a_2(x)^3+a_3(x)^3 = x$ 
$a_1(x)^5+a_2(x)^5+a_3(x)^5+...+a_6(x)^5 = x$
$a_1(x)^7+a_2(x)^7+a_3(x)^7+a_4(x)^7+...+a_8(x)^7 = x$
where the given number of addends $a_i$ are rational functions in terms of x, hence can be treated as "Waring-like problems". The case k = 3 mentioned by Esteban is cited in Yuri Manin's book Cubic Forms (but has been solved earlier by Ryley), while k = 5 and 7 have been solved by Choudhry. Interestingly, k = 7 involves polynomials where the coefficients have 33 digits! One has to solve the simultaneous equations,
$a^2+m^5b^2 = c^2+m^5d^2$
$a^4+m^3b^4 = c^4+m^3d^4$
which, for m = 2, can be reduced to an elliptic curve and has an infinite number of integer solutions (after multiplying out the denominators), though the "smallest" one has 33 digits.
P.S. If anybody can solve k = 9, I'll be interested to know.
See, "Waring-Like Problems" at http://sites.google.com/site/tpiezas/001b.
A: Here is a start:
As ihf observed, you need
$f(x)^3+g(x)^3+h(x)^3=x q(x)^3$
On the left side you have three polinomials of degrees multiple of 3, on the right the degree is M3+1, thus something must cancel on the left side.
Without loss of generality, you can consider 2 cases:
Case 1: $deg(f)=deg(g) = deg(h)$. In this case, let $ax^k, bx^k$ and $cx^k$ be the dominant monomials in $f,g,h$. Then 
$$a^n+b^n+c^n=0 \,.$$
which is not possible by Fermat last Theorem. So this is not possible.
Case 2: $deg(f)=deg(g) \geq deg(h)+1$. 
Then $f=ax^k+bx^{k-1}+...$.  In order for the first two terms to cancel you need 
$g=-ax^k-bx^{k-1}+..$.  And I am stucked here.
This suggest that the simplest potential example would be something like:
$$f=x^2+ax+b \,;\, g=-x^2-ax+c \,;\, h(x)=dx+e \,.$$
The computations seem too long :)
A: To the question added later:
$f(x)^3 + g(x)^3 = x$
It is not possible.
Setting $x$ to be $1^3, 2^3, \dots $ gives us a non-trivial solution to the diophantine equation $x^3 + y^3 = z^3$, since $f$ and $g$ can only have finitely many roots.
