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How to prove that every polynomial with real coefficients is the sum of three polynomials raised to the 3rd degree? Formally the statement is:

$\forall f\in\mathbb{R}[x]\quad \exists g,h,p\in\mathbb{R}[x]\quad f=g^3+h^3+p^3$

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    $\begingroup$ I'm confused by this question. Surely a linear function or a quadratic for $f$ is a simple counter example. $\endgroup$
    – Ian Miller
    Oct 30, 2016 at 8:24
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    $\begingroup$ @IanMiller No. $3x^2+3x=(x+1)^3+(-x)^3+(-1)^3.$ $\endgroup$
    – mfl
    Oct 30, 2016 at 8:29
  • $\begingroup$ Oh right. My bad. $\endgroup$
    – Ian Miller
    Oct 30, 2016 at 9:25
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    $\begingroup$ What is this, a contest question or something? $\endgroup$
    – Jack M
    Oct 30, 2016 at 23:11
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    $\begingroup$ @Glinka what contest? $\endgroup$
    – frog1944
    Nov 12, 2016 at 1:36

1 Answer 1

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We have that the following identity holds $$(x+1)^3+2(-x)^3+(x-1)^3=6x.$$ Hence $$\left(\frac{f(x)+1}{6^{1/3}}\right)^{3}+\left(\frac{-f(x)}{3^{1/3}}\right)^{3}+ \left(\frac{f(x)-1}{6^{1/3}}\right)^{3}=f(x).$$

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    $\begingroup$ Great answer. This appears to be extendable to all powers as: $$\sum_{i=0}^{n-1}\left(x-\frac{n-1}{2}+i\right)^n\cdot(-1)^i\cdot{n-1\choose i}=n!\cdot x$$ $\endgroup$
    – Ian Miller
    Oct 30, 2016 at 10:08
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    $\begingroup$ @IanMiller For $n = 2$ you get $-2x$ on the left. I think the right side should be $(-1)^{n-1} n! \cdot x$ $\endgroup$
    – Jay
    Oct 31, 2016 at 0:15
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    $\begingroup$ @Jay Thanks. I did drop that bit somehow from my expression. $\endgroup$
    – Ian Miller
    Oct 31, 2016 at 3:59
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    $\begingroup$ Over the rationals: I wonder for what $f\in\mathbb{Q}[x]$ there exist $g,h,p\in\mathbb{Q}[x]$ such that $f=g^3+h^3+p^3$ (since the cube root of six is clearly irrational). $\endgroup$ Oct 31, 2016 at 10:20
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    $\begingroup$ @JeppeStigNielsen : this could be related. $\endgroup$
    – Watson
    Oct 31, 2016 at 20:23

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