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 $1+\omega+\omega^2=0$ and $\omega^3=1$. Expanding the expression inside the summation:

Vedclass · Accepted Answer

$285$

Vedclass · Answer

(A) Given that $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2=0$ and $\omega^3=1$. Expanding the expression inside the summation: $(\omega x+2)(\omega^2 x+2)-3 = \omega^3 x^2 + 2\omega x + 2\omega^2 x + 4 - 3$ $= x^2 + 2x(\omega + \omega^2) + 1$ (since $\omega^3=1$) $= x^2 + 2x(-1) + 1 = x^2 - 2x + 1 = (x-1)^2$. Now,we calculate the sum: $\sum_{x=1}^{10} (x-1)^2 = \sum_{x=1}^{10} (x^2 - 2x + 1) = \sum_{x=1}^{10} x^2 - 2\sum_{x=1}^{10} x + \sum_{x=1}^{10} 1$ $= \frac{10(11)(21)}{6} - 2 	imes \frac{10(11)}{2} + 10$ $= 385 - 110 + 10 = 285$.

If $\omega$ is a complex cube root of unity,then find the value of $\sum_{x=1}^{10} ((\omega x+2)(\omega^2 x+2)-3)$.

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