Question

If 1,w and w² are cubic roots of 1 ,then (1-w)(1-w²)(1-w⁴)(1-w⁸)=

Answer 1

We have:

$(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^8)$

We have to find, the value of $(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^8)$ is:

Solution:

∴ $(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^8)$

= $(1-\omega)(1-\omega^2)(1-\omega^3.\omega)(1-(\omega^3)^2\omega^2)$

= $(1-\omega)(1-\omega^2)(1-\omega)(1-\omega^2)$ [ ∵ $\omega^3=1$ ]

= $(1-\omega)^2(1-\omega^2)^2$

= $(1-2\omega+\omega^2)(1-2\omega^2+\omega^4)$

= $(1-2\omega+\omega^2)(1-2\omega^2+\omega)$ [ ∵ $\omega^3=1$ ]

= $(1-2\omega+\omega^2)(1-2\omega^2+\omega)$

= $(-2\omega-\omega)(-2\omega^2-\omega^2)$ [ ∵ $1+\omega+\omega^2=0$ ]

= $(-3\omega)(-3\omega^2)$

= $9\omega^3$

= 9

∴ $(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^8)$ = 9

Answer 2

Given : 1, w w² is cube root of 1

To Find : (1-w)(1-w²)(1-w⁴)(1-w⁸)

Solution:

1, w w² is cube root of 1

=> x³ = 1

=> x³ - 1  = 0

1 + w + w² = -0/1 = 0

1.w.w² = w³ = -(-1)/1 = 1

1 + w + w² =  0  =>   w + w² = - 1

w³  = 1

(1-w)(1-w²)(1-w⁴)(1-w⁸)

w⁴ = w³.w = w

w⁸ = w³.w³.w² = w²

= (1-w)(1-w²)(1-w)(1-w²)

= ((1 - w)(1 - w²))²

= (1 - w - w² + w³)²

= ( 1 -(w + w²) + w³)²

w + w² = - 1  , w³= 1

= ( 1 - (-1) + 1)²

= 3²

= 9

(1-w)(1-w²)(1-w⁴)(1-w⁸)  = 9

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