Question

Statement 1: If n is an odd integer greater than 3 but not a multiple of 3, then $(x+1{)}^n-x^n-1$ is divisible by $x^3+x^2+x$.
Statement 2: If n is an odd integer greater than 3 but not a multiple of 3, we have $1+{\omega }^n+{\omega }^{2n}=3$.

• A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.
• B. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.
• C. Statement 1 is true and Statement 2 is false.
• D. Statement 1 is false and Statement 2 is true.

Answer 1

Answer: (C)

Statement 1 is true and Statement 2 is false.

$x^3+x^2+x=x(x^2+x+1)=x(x-\omega )(x-{\omega }^2)$
Now $f\left(x\right)=(x+1{)}^n-x^n-1$ is divisible by $x^3+x^2+x$. Then
$f\left(0\right)=0,f\left(\omega \right)=0f\left({\omega }^2\right)=0$. Now,
$f\left(0\right)=(0+1{)}^n-0^n-1=0$
$f\left(\omega \right)=(\omega +1{)}^n-{\omega }^n-1$
$=(-{\omega }^2{)}^n-{\omega }^n-1=-({\omega }^{2n}+{\omega }^n+1)=0$ (as n is not a multiple of 3)
Similarly, we have $f\left({\omega }^2\right)=0$.
Hence, statement 1 is correct but statement 2 is false.