Question

$x^{3m}+x^{3n-1}+x^{3r-2}$, where $m,n,rϵN$ is divisible by

• A. $x^2-x+1$
• B. $x^2+x+1$
• C. $x^2+x-1$
• D. $x^2-x-1$

Answer 1

The correct option is B $x^2+x+1$

Consider the eq. $1+{x+x}^2=0$
Roots of the above eq. are $w$ & $w^2$
where $w$ is the cube root of unity.
$\therefore w^3=1\&1+{w+w}^2=0$
$f\left(x\right)=x^{3m}+x^{3n-1}+x^{3r-2}$
$f\left(w\right)=w^{3m}+w^{3n-1}+w^{3r-2}={\left(w^3\right)}^m+\frac{{\left(w^3\right)}^n}{w}+\frac{{\left(w^3\right)}^r}{w^2}\Rightarrow f\left(w\right)=\frac{1+{w+w}^2}{w^3}=0$
$f\left(w^2\right)=w^{6m}+w^{6n-2}+w^{6r-4}={\left(w^3\right)}^{2m}+\frac{{\left(w^3\right)}^{2n}}{w^2}+\frac{{\left(w^3\right)}^{2r}}{w^4}\Rightarrow f\left(w^2\right)=\frac{1+w^4{+w}^2}{w^6}=1+{w+w}^2=0$
Since ${w\&w}^2$ are the roots of the equation $f\left(x\right)=0$
$\Rightarrow f\left(x\right)$ is divisible by $(x-w)(x-w^2)$
Therefore $f\left(x\right)$ is divisible by $1+{x+x}^2$.
Hence, option 'B' is correct.