2016/082) Let P(x) be a polynomial of x. Show that there exists a polynomial Q(x) such that P(x)Q(x) is a polynomial of $x^3$

We have
$P(x) = \sum_{n=0}^{\infty} a_nx^n$
this can be written as $A(x) + x B(x) + x^2C(x)$
where
$A(x) = \sum_{k=0}^{\infty} a_{3n}x^{3n}$
$B(x) = \sum_{k=0}^{\infty} a_{3n+1}x^{3n}$
$C(x) = \sum_{k=0}^{\infty} a_{3n+2}x^{3n}$
Now let
$R(x) =  A(x) + x B(x)w + x^2C(x)w^2$ where w is cube root of one
and
$S(x) = A(x) + x B(x)w^2 + x^2C(x)w$
using $(a+b+c)(a+bw^2+cw)(a+bw+cw^2) = a^3+b^3+c^3 - 3abc$
we get $P(x)R(x)S(x) = (\sum_{k=0}^{\infty} a_{3n}x^{3n})^3 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^3 x^3 + (\sum_{k=0}^{\infty} a_{3n+2}x^{3n})^3x^6$
$- 3(\sum_{k=0}^{\infty} a_{3n}x^{3n})(\sum_{k=0}^{\infty} a_{3n+1}x^{3n})(\sum_{k=0}^{\infty} a_{3n+2}x^{3n})x^3$
which is a polynomial of $x^3$
now using  $(a+bw+cw^2)( a+ bw^2 +cw) = (a^2+b^2+c^2-ab-bc-ca)$
we have
$R(x)S(x) = (A(x) + x B(x)w + x^2C(x)w^2)(A(x) + x B(x)w^2 + x^2C(x)w)$
$= (\sum_{k=0}^{\infty} a_{3n}x^{3n})^2 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^2)x^2 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^2x^4$
   $- (\sum_{k=0}^{\infty} a_{3n+1}x^{3n}) (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})x$
  $- (\sum_{k=0}^{\infty} a_{3n+1}x^{3n}))(\sum_{k=0}^{\infty} a_{3n+2}x^{3n})x^3$
  $-(\sum_{k=0}^{\infty} a_{3n+2}x^{3n}\sum_{k=0}^{\infty} a_{3n}x^{3n})x^3$
multiplying by the above polynomial say Q(x) we get a polynomial of $x^3$