Question

If $(x+1{)}^n-x^n-1$ is divisible by $x^3+x^2+x$ then which of the following is true :

• A. $n$ is not a multiple of $3$ and odd number
• B. $n$ is a multiple of $3$ and odd number
• C. $n$ is a multiple of $3$ and even number
• D. $n$ is not a multiple of $3$ and even number

Answer 1

The correct option is A $n$ is not a multiple of $3$ and odd number

Let
$f\left(x\right)=(x+1{)}^n-x^n-1$
As $x^3+x^2+x=x(x^2+x+1)=x(x-\omega )(x-{\omega }^2)$ is factor of $f\left(x\right)$,
$\Rightarrow f\left(0\right)=0,f\left(\omega \right)=0,f\left({\omega }^2\right)=0$
Now $\left(i\right)$$f\left(0\right)=(0+1{)}^n-0^n-1=0$
$f\left(0\right)=0\forall n\in Z$

$\left(ii\right)f\left(\omega \right)=(\omega +1{)}^n-{\omega }^n-1=(-{\omega }^2{)}^n-{\omega }^n-1$
Now, $f\left(\omega \right)=0$ iff $n$ is not multiple of $3$ and odd number, as $(-{\omega }^2{)}^n-{\omega }^n-1=-({\omega }^{2n}+{\omega }^n+1)=0$

$\left(iii\right)f\left({\omega }^2\right)=({\omega }^2+1{)}^n-{\omega }^{2n}-1=(-\omega {)}^n-{\omega }^{2n}-1)$
Now, $f\left({\omega }^2\right)=0$ iff $n$ is not multiple of $3$ and odd number, as $(-\omega {)}^n-{\omega }^{2n}-1=-({\omega }^n+{\omega }^{2n}+1)=0$
$\therefore n$ should not be a multiple of $3$ and $n$ is odd number.