Question

Question :


If $1,\omega,\omega^2$ are the three cube roots of unity , then $(3+\omega^2+\omega^4)^6)$ = ?​

Answer 1

Solution:

( 3 + 3ω² + 5ω)⁶ - ( 2+ 6ω² + 2ω)³

= ( 3+3ω² + 3ω+ 2ω)⁶ - (2+ 4ω² + 2ω² + 2ω)³

= (3 (1 + w² + w) + 2ω)⁶ - (2(1 +ω² + ω) + 4ω

= (2ω⁶) - (4ω²)³ = 0 ( 1 + ω + ω²=0)

Answer is 0

Answer 2

$\huge\bf{Question:-}$

$</p><p>If \: 1,\omega,\omega^2$ are the three cube roots of unity , then $(3+\omega^2+\omega^4)^6) = ?$

$\huge\bf{Solution}$

• (Refer to attachment)

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