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Cube Root of a Complex Number

If ω is a com...

Question

If $ω$ is a complex cube root of unity, then the value of $(a + b)^{2} + (a ω + b ω^{2})^{2} + (a ω^{2} + b ω)^{2}$ is

A

6 a b

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B

3 a b

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C

12 a b

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D

a b

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Solution

The correct option is A $6 a b$
The given expression is -
$(a + b)^{2} + (a ω + b ω^{2})^{2} + (a ω^{2} + b ω)^{2}$
$= (a^{2} + b^{2} + 2 a b) + (a^{2} ω^{2} + b^{2} ω^{4} + 2 a b ω^{3}) + (a^{2} ω^{4} + b^{2} ω^{2} + 2 a b ω^{3})$

$= (a^{2} + b^{2} + 2 a b) + (a^{2} ω^{2} + b^{2} ω + 2 a b)$
$+ (a^{2} ω + b^{2} ω^{2} + 2 a b)$
$[∵ ω^{3} = 1]$

$= a^{2} (1 + ω + ω^{2}) + b^{2} (1 + ω + ω^{2}) + 6 a b$
$= 6 a b (∵ 1 + ω + ω^{2} = 0)$

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Cube Root of a Complex Number

Standard XII Mathematics

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