If 1, omega, omega^2 are the cube roots of un | vedclass

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(C) Given that $(1+\omega)$ is a root of $x^n-x=0$. Substituting $x = 1+\omega$,we get $(1+\omega)^n - (1+\omega) = 0$. Since $1+\omega+\omega^2=0$,we

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Given that $(1+\omega)$ is a root of $x^n-x=0$. Substituting $x = 1+\omega$,we get $(1+\omega)^n - (1+\omega) = 0$. Since $1+\omega+\omega^2=0$,we have $1+\omega = -\omega^2$. Substituting this into the equation: $(-\omega^2)^n - (-\omega^2) = 0$. $(-1)^n \omega^{2n} + \omega^2 = 0$. $(-1)^n \omega^{2n} = -\omega^2$. For this to hold,$n$ must be odd (so $(-1)^n = -1$) and $\omega^{2n} = \omega^2$. This implies $2n \equiv 2 \pmod{3}$,which means $2n = 3k + 2$ for some integer $k$. For $n=3$,$2(3) = 6 \equiv 0 \pmod{3}$ (False). For $n=5$,$2(5) = 10 \equiv 1 \pmod{3}$ (False). For $n=7$,$2(7) = 14 = 3(4) + 2 \equiv 2 \pmod{3}$ (True). Thus,the least value of $n$ is $7$.

If $1, \omega, \omega^2$ are the cube roots of unity,$n \in \mathbb{N}$ and $n > 2$,then the least value of $n$ such that $1+\omega$ is a root of $x^n-x=0$ is

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