If omega is a complex cube root of unity,then | vedclass

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(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Consider the expression $(x+1)(x+\omega

Vedclass · Accepted Answer

$x^3-1$

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(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Consider the expression $(x+1)(x+\omega)(x-\omega-1)$.Note that $-\omega-1 = \omega^2$ since $1 + \omega + \omega^2 = 0$.So,the expression becomes $(x+1)(x+\omega)(x+\omega^2)$.We know the identity $(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc$.Here,the roots are $-1, -\omega, -\omega^2$.Thus,$(x+1)(x+\omega)(x+\omega^2) = x^3 - (-1-\omega-\omega^2)x^2 + (\omega+\omega^2+\omega^3)x - (1)(\omega)(\omega^2)$.Since $1+\omega+\omega^2=0$,we have $-1-\omega-\omega^2=0$.Also,$\omega+\omega^2+\omega^3 = -1 + 1 = 0$.And $\omega^3 = 1$.Therefore,the expression simplifies to $x^3 - (0)x^2 + (0)x - 1 = x^3 - 1$.

If $\omega$ is a complex cube root of unity,then $(x+1)(x+\omega)(x-\omega-1)$ is equal to

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