Question

9. If w is the complex cube root of unity show that (1 + w – w2)6 = 64
​

Answer 1

Answer:

( 1 + ω - ω² )⁶ = 64

Step-by-step-explanation:

We have given that,

ω is the complex cube root of unity.

We have to show that,

( 1 + ω - ω² )⁶ = 64

Now, we know that,

If ω is a complex cube root of unity,

ω³ = 1

1 + ω + ω² = 0

∴ 1 + ω = - ω²

Now, considering the LHS of the given equation,

( 1 + ω - ω² )⁶ = 64

LHS = ( 1 + ω - ω² )⁶

⇒ LHS = ( - ω² - ω² )⁶

⇒ LHS = ( - 2 ω² )⁶

⇒ LHS = ( - 2 )⁶ * ( ω² )⁶

⇒ LHS = 64 * ( ω )⁽ ⁶ * ² ⁾

⇒ LHS = 64 * ( ω )¹²

⇒ LHS = 64 * ( ω )⁽ ⁴ * ³ ⁾

⇒ LHS = 64 * ( ω³ )⁴

⇒ LHS = 64 * ( 1 )⁴

⇒ LHS = 64 * 1

⇒ LHS = 64

RHS = 64

∴ LHS = RHS

Hence shown!

Answer 2

Question:-

If ω is the complex cube root of unity show that

(1 + ω – ω²)⁶ = 64.

Given:-

• (1 + ω – ω²)⁶ = 64.

To Find:-

• Need to Find that (1 + ω – ω²)⁶ = 64.

Solution:-

ω is the complex cube root of unity.

∴ ω³ = 1 and 1 + ω + ω² = 0.

Also, 1 + ω² = – ω, 1 + ω = – ω² and ω + ω² = – 1.

L.H.S. = (1 + ω – ω²)⁶ .

= [ (1 + ω) – ω² ]⁶.

= ( – ω² – ω² )⁶.

= ( – 2ω² )⁶.

= 64.ω¹² .

= 64(ω³)⁴.

= 64(1)⁴.

= 64.

Answer:-

$\sf \large{ \boxed{ \sf \large \red{(1 + ω – ω²)⁶ = 64.}}} \\ \\ \sf \large{ \boxed{ \sf \large \blue{Hence, \: Proved \: that = L.H.S. = R.H.S.}}}$

Hope you have satisfied. ⚘