(A) We know that for a cube root of unity,$1 + \omega + \omega^2 = 0$. Therefore,$1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$. Substituting t

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(A) We know that for a cube root of unity,$1 + \omega + \omega^2 = 0$. Therefore,$1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$. Substituting these values into the expression: $(1 + \omega)^3 - (1 + \omega^2)^3 = (-\omega^2)^3 - (-\omega)^3$ $= -\omega^6 - (-\omega^3)$ $= -(\omega^3)^2 + \omega^3$ Since $\omega^3 = 1$,we have: $= -(1)^2 + 1 = -1 + 1 = 0$.

Class 11 Mathematics 4-1.Complex numbers De Moivre's theorem and Roots of unity

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If $\omega$ is a cube root of unity,then $(1 + \omega)^3 - (1 + \omega^2)^3 = $

A
$0$
B
$\omega$
C
$\omega^2$
D
None of these

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4-1.Complex numbers

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