Question 917729
{{{P(x)=g(x^3)+x*h(x^3)}}}
{{{g(x)=(x-1)G(x)+r}}} for some quotient polynomial {{{G(x)}}} and some number remainder {{{r}}} .
(If {{{r=0}}}, then {{{g(x)}}} is divisible {{{(x-1)}}} ).
{{{h(x)=(x-1)H(x)+s}}} for some quotient polynomial {{{H(x)}}} and some number remainder {{{s}}} .
(If {{{s=0}}}, then {{{h(x)}}} is divisible {{{(x-1)}}} ).
 
Substituting, we get
{{{P(x)=(x^3-1)G(x^3)+r+x*((x^3-1)H(x^3)+s)}}}
{{{P(x)=(x^3-1)G(x^3)+r+x*(x^3-1)H(x^3)+sx}}}
{{{P(x)=(x^3-1)(G(x^3)+x*H(x^3))+sx+r}}}
Since {{{x^3-1=(x-1)(x^2+x+1)}}} ,
{{{P(x)=(x^2+x+1)(x-1)(G(x^3)+x*H(x^3))+sx+r}}}
That means that
when you divide {{{P(x)}}} by the quadratic polynomial {{{(x^2+x+1)}}} ,
the quotient is {{{(x-1)(G(x^3)+x*H(x^3))}}} ,
and the remainder is the linear polynomial {{{rx+s}}} .
Since {{{P(x)}}} is divisible by {{{(x^2+x+1)}}} ,
{{{rx+s}}} must be zero for all values of {{{x}}} , meaning that {{{r=s=0}}} .