Question

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

Answer 1

$\Large{\underbrace{\sf{\red{Required\:Solution:}}}}$

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

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

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

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

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

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

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

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

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

= $9\omega^3$

= 9

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

___________________________

Answer 2

Step-by-step explanation:

RequiredSolution:

∴ (1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^8)(1−ω)(1−ω

2

)(1−ω

4

)(1−ω

8

)

= (1-\omega)(1-\omega^2)(1-\omega^3.\omega)(1-(\omega^3)^2\omega^2)(1−ω)(1−ω

2

)(1−ω

3

.ω)(1−(ω

3

)

2

ω

2

)

= (1-\omega)(1-\omega^2)(1-\omega)(1-\omega^2)(1−ω)(1−ω

2

)(1−ω)(1−ω

2

) [ ∵ \omega^3=1ω

3

=1 ]

= (1-\omega)^2(1-\omega^2)^2(1−ω)

2

(1−ω

2

)

2

= (1-2\omega+\omega^2)(1-2\omega^2+\omega^4)(1−2ω+ω

2

)(1−2ω

2

+ω

4

)

= (1-2\omega+\omega^2)(1-2\omega^2+\omega)(1−2ω+ω

2

)(1−2ω

2

+ω) [ ∵ \omega^3=1ω

3

=1 ]

= (1-2\omega+\omega^2)(1-2\omega^2+\omega)(1−2ω+ω

2

)(1−2ω

2

+ω)

= (-2\omega-\omega)(-2\omega^2-\omega^2)(−2ω−ω)(−2ω

2

−ω

2

) [ ∵ 1+\omega+\omega^2=01+ω+ω

2

=0 ]

= (-3\omega)(-3\omega^2)(−3ω)(−3ω

2

)

= 9\omega^39ω

3

= 9

∴ (1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^8)(1−ω)(1−ω

2

)(1−ω

4

)(1−ω

8

) = 9

___________________________