Question

If $A\left(x\right)$ and $B\left(x\right)$ be two polynomials and $f\left(x\right)=A\left(x^3\right)+xB\left(x^3\right)$. If $f\left(x\right)$ is divisible by $x^2+x+1$ then show that it is divisible by $x-1$ also.

Answer 1

Given $f\left(x\right)=A\left(x^3\right)+xB\left(x^3\right)$ is divisible by $x^2+x+1=(x-\omega )(x-{\omega }^2)$ where ${\omega }^3=1$.

This gives $f\left(x\right)$ is also divisible by $(x-\omega )$ and $(x-{\omega }^2)$.

So we have $f\left(\omega \right)=A\left(1\right)+\omega B\left(1\right)=0$.......(1) and $f\left({\omega }^2\right)=A\left(1\right)+{\omega }^{2}B\left(1\right)=0$.....(2). [ Since ${\omega }^3=1$]

Solving from (1) and (2) we get $A\left(1\right)=0=B\left(1\right)$.

Now $f\left(1\right)=A\left(1\right)+B\left(1\right)=0$ [ Using the values of $A\left(1\right)$ and $B\left(1\right)$.]

So using remainder theorem we have $f\left(x\right)$ is also divisible by $(x-1)$.