If omega neq 1 is a cube root of unity and a - b - 21- a-3 - b-3 - 8001- then the value of dfrac-aomega-2- -bomega -aomega -bomega-2- -127- must be equal to1034

Question

Toppr Admin · Accepted Answer

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If $ω \neq 1$ is a cube root of unity and $a + b = 21, a^{3} + b^{3} = 8001$ , then the value of $\frac{(a ω^{2} + b ω) (a ω + b ω^{2})}{127}$ must be equal to
$1$
$0$
$3$
$4$

$(a ω^{2} + b ω) (a ω + b ω^{2}) = a^{2} - a b + b^{2}$

If ω≠1 is a cube root of unity and a+b=21,a3+b3=8001, then the value of (aω2+bω)(aω+bω2)127 must be equal to1034

(aω2+bω)(aω+bω2)=a2−ab+b2

If $ω \neq 1$ is a cube root of unity and $a + b = 21, a^{3} + b^{3} = 8001$ , then the value of $\frac{(a ω^{2} + b ω) (a ω + b ω^{2})}{127}$ must be equal to
$1$
$0$
$3$
$4$

$(a ω^{2} + b ω) (a ω + b ω^{2}) = a^{2} - a b + b^{2}$