Complex Numbers 5 Question 18
19.
It is given that $n$ is an odd integer greater than 3 , but $n$ is not a multiple of 3 . Prove that $x^{3}+x^{2}+x$ is a factor of $\quad(x+1)^{n}-x^{n}-1$
(1980, 3M)
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Solution:
- Since, $n$ is not a multiple of 3 , but odd integers and
$ x^{3}+x^{2}+x=0 \Rightarrow x=0, \omega, \omega^{2} $
Now, when $x=0$
$\Rightarrow(x+1)^{n}-x^{n}-1=1-0-1=0$
$\therefore \quad x=0$ is root of $(x+1)^{n}-x^{n}-1$
Again, when $x=\omega$
$\Rightarrow(x+1)^{n}-x^{n}-1=(1+\omega)^{n}-\omega^{n}-1=-\omega^{2 n}-\omega^{n}-1=0$
[as $n$ is not a multiple of 3 and odd]
Similarly, $x=\omega^{2}$ is root of ${(x+1)^{n}-x^{n}-1 }$
Hence, $x=0, \omega, \omega^{2}$ are the roots of $(x+1)^{n}-x^{n}-1$
Thus, $x^{3}+x^{2}+x$ divides $(x+1)^{n}-x^{n}-1$.
