For a): $ 3x+3x^7=-x^9+(x+x^3)^3-(x+x^2)^3+(x^2+1)^3+(x-1)^3-x^3$. For b): Apply the lemma from here, the second post to get the negative answer.

Can you say me who is the proposer? Because this exact problem can be found at http://reflections.awesomemath.org/reflections/2008_6/MR_6_2008_problems.pdf see O106 Proposed by Nairi Sedrakian, Yerevan, Armenia!

Well, I think the author is the same Nairi Sedrakian .

simply because he was and to my knowledge, is going to be again in the jury of ZIMO 2009.

i think there is a little simpler solution for the second one we have that $ (\sum_{i = 0}^n a_ix{}^i){}^3 = \sum_{i = 0}^na_i{}^3x{}^{3i} + 3\sum_{0\leq i < j\leq n}a_i^2a_jx^{2i + j} + a{}_ia{}_j{}^2x{}^{i + 2j} + 6\sum_{0\leq i < j < k\leq n}a{}_ia{}_j{}_a{}_k$ this implies that the sum of the coefficients of any "non-cube" term is divisible by $ 6$, since $ a_i^2a_k + a{}_ia{}_j^2$ is always even, and we're done

what? $ i+2j\neq 2i+j$. That invoked misconception, I suspect.

i mean that $ i + 2j$ is not congruent to $ 0$ iff $ 2i + j$ is not congruent to $ 0$, so $ x^{i + 2j}$ is a "non-cube", iff $ x^{2i + j}$ is a "non-cube"... i just realized that i wrote something wrong, i wanted to say that "the sum of the coefficients of all "non-cube" terms is divisible by $ 6$"... so, for any combination of cubes of polynomials we would have that the sum of the coefficients of "non-cube" terms will be divisible by $ 6$, which is not the case of $ 3 + 3 + 3$... is it clear now?

I'd like to solve only a here,for b is trivial. it's easy to prove that if f and g are good then f+g and f-g are good by expanding $(x^2+x+1)^3-(x^2+x)^3$ we know that $3x^4+3x$ is good replacing $x$ with $-x$,then $3x^4-3x$ is good. so $3x^7+3x=3x^7+3x^4-(3x^4-3x)=x^3(3x^4+3x)-(3x^4-3x)$ is good.

We will prove that the sum of of the coefficients near the exponent of the form $3k+1,3k+2$. This follows from: $$P(\omega)=b_0+b_1\omega+b_2\omega^2\equiv_{2} \sum (x+y)(x+z)(z+y)$$But $(x+y)(y+z)(z+x)$ is even ,as by php there are two numbers with the same modulo class 2. This kills b). For a) it is possible ,namely: $$(1+x+x^3)^3+(-x^2-1)^3+(-x^2-x)^3+4(-x)^3+(-x^2)^3+(-x^3)^3=(3x+3x^7)$$

For a), simply play around with cubes of polynomials of the form $\varepsilon_{3}x^3 + \varepsilon_{2}x^2 + \varepsilon_{1}x + \varepsilon_{0}$ for $\varepsilon_{i} \in \{-1, 0, 1\}$ to get a construction. For example: \[(x^3+x+1)^3 + (-x^3)^3 + (-x^2-x-1)^3 + (-x^2)^3 + (-x^2)^3 + (x+1)^3 + x^3 + x^3 + (-1)^3 = 3x^7 + 3x\]For b), we work over $\mathbb{Z}[x]/(x^3-1)$. The polynomial we want to express becomes $9x$, but if we assume there exist reduced quadratic polynomials such that $9x = \sum\limits_{i=1}^{k} p_{i}^3$, then after writing $p_{i}(x) = a_{i}x^2 + b_{i}x^2 + c_{i}$, the sum becomes \[\sum\limits_{i=1}^{k} x^2(3a_{i}^2 b_{i}+3b_{i}^2c_{i}+3c_{i}^2a_{i})+x(3a_{i}^2c_{i}+3c_{i}^2b_{i}+3b_{i}^2a_{i})+a_{i}^3+b_{i}^3+c_{i}^3+6a_{i}b_{i}c_{i}\]Here $3a_{i}^2 b_{i}+3b_{i}^2c_{i}+3c_{i}^2a_{i} \equiv 3a_{i}^2c_{i}+3c_{i}^2b_{i}+3b_{i}^2a_{i} \pmod 2$ for all $i$, which doesn't hold for the polynomial $9x$, so we reach a contradiction.

also recent elmo problem