Every polynomial with real coefficients is the sum of cubes of three polynomials
Solution 1:
We have that the following identity holds $$(x+1)^3+2(-x)^3+(x-1)^3=6x.$$ Hence $$\left(\frac{f(x)+1}{6^{1/3}}\right)^{3}+\left(\frac{-f(x)}{3^{1/3}}\right)^{3}+ \left(\frac{f(x)-1}{6^{1/3}}\right)^{3}=f(x).$$