Question

If $\omega \ne 1$ is a cube root of unity and ${(\omega +x)}^n=1+12\omega +69\omega +....$ then values of $4n$ and $\omega$ respectively are

• A. $36,1$
• B. $12,2$
• C. $24,1/2$
• D. $18,1/3$

Answer 1

The correct option is C $24,1/2$

From the above expansion, we can conclude
$w^n=1$
$\Rightarrow n.w^{n-1}.x=\frac{nx}{w}=12w$
and
$\frac{n(n-1)w^{n-2}{x}^2}{2}=\frac{x^{2}n(n-1)}{2w^2}=69w$
Now
$w^n=1$ hence n is multiple of 3..(i)
And
Therefore
Comparing coefficients, we get
$nx=12$
$\Rightarrow \frac{n(n-1)x^2}{2}=69$
$\Rightarrow n(n-1)x^2=138$
$\Rightarrow nx(nx-x)=138$
$\Rightarrow 12(12-x)=138$
$\Rightarrow 144-12x=138$
$\Rightarrow 6=12x$
$\Rightarrow x=0.5$
We Know that $|w|=1$
Also
$nx=12$
$\Rightarrow n=12.\left(2\right)=24$