Let omega be a cube root of unity not equal t | vedclass

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(B) We are given the expression $|a + b\omega + c\omega^2|$ where $a, b, c \in \{1, -1\}$.Since $\omega$ is a cube root of unity,we know that $1 + \om

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$2$

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(B) We are given the expression $|a + b\omega + c\omega^2|$ where $a, b, c \in \{1, -1\}$.Since $\omega$ is a cube root of unity,we know that $1 + \omega + \omega^2 = 0$,which implies $\omega + \omega^2 = -1$.We test the possible combinations of $a, b, c \in \{1, -1\}$:If $a=1, b=-1, c=-1$,then $|1 - \omega - \omega^2| = |1 - (\omega + \omega^2)| = |1 - (-1)| = |1 + 1| = 2$.If $a=1, b=1, c=-1$,then $|1 + \omega - \omega^2| = |1 + \omega - (-1 - \omega)| = |2 + 2\omega| = 2|1 + \omega| = 2|-\omega^2| = 2$.If $a=1, b=1, c=1$,then $|1 + \omega + \omega^2| = |0| = 0$.Thus,the maximum possible value is $2$.

Let $\omega$ be a cube root of unity not equal to $1$. Then,the maximum possible value of $|a + b\omega + c\omega^2|$,where $a, b, c \in \{+1, -1\}$ is

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