# Every polynomial with real coefficients is the sum of cubes of three polynomials

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

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

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

We have that the following identity holds $$(x+1)^3+2(-x)^3+(x-1)^3=6x.$$ Hence $$\left(\frac{f(x)+1}{6^{1/3}}\right)^{3}+\left(\frac{-f(x)}{3^{1/3}}\right)^{3}+ \left(\frac{f(x)-1}{6^{1/3}}\right)^{3}=f(x).$$
• Great answer. This appears to be extendable to all powers as: $$\sum_{i=0}^{n-1}\left(x-\frac{n-1}{2}+i\right)^n\cdot(-1)^i\cdot{n-1\choose i}=n!\cdot x$$ Oct 30, 2016 at 10:08
• @IanMiller For $n = 2$ you get $-2x$ on the left. I think the right side should be $(-1)^{n-1} n! \cdot x$
• 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). Oct 31, 2016 at 10:20