If 1- - and -2 are cube roots of unity- show that 1-1-21-41-8-1-

Given, (1+w)(1+w^2)(1+w^4)(1+w^8) considering first 2 terms, we get, #160; (1+w)(1+w^2)=1+w^2+w+w^3=0+w^3=1 similarly, considering second 2 terms, we ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Discriminant

If 1, ω and...

Question

If $1, ω$ and $ω^{2}$ are cube roots of unity, show that $(1 + ω) (1 + ω^{2}) (1 + ω^{4}) (1 + ω^{8}) = 1$ .

Open in App

Solution

Given,

$(1 + w) (1 + w^{2}) (1 + w^{4}) (1 + w^{8})$

considering first 2 terms, we get,

$(1 + w) (1 + w^{2}) = 1 + w^{2} + w + w^{3} = 0 + w^{3} = 1$

similarly, considering second 2 terms, we get,

$(1 + w^{4}) (1 + w^{8}) = (1 + w) (1 + w^{2}) = 1$

$1 \times 1 = 1$

Hence proved.

Suggest Corrections

Similar questions

Q. If

1, ω

and

ω^{2}

are cube roots of unity, show that

(1 - ω) (1 - ω^{2}) (1 - ω^{4}) (1 - ω^{8}) = 9

Q. If

1, ω

and

ω^{2}

are cube roots of unity, show that

(1 + ω) (1 + ω^{2}) (1 + ω^{4}) (1 + ω^{8})

upto

2

n factors

= 1

Q. If

1, ω, ω^{2}

are three cube roots of unity, then

(1 - ω + ω^{2}) (1 + ω - ω^{2})

Q. lf

1, ω, ω^{2}

are cube roots of unity then the value

(1 - ω + ω^{2}) (1 - ω^{2} + ω^{4}) (1 - ω^{4} + ω^{8}) \dots . . .

(upto 2n factors) is?

Q. If

1, ω, ω^{2}

are the three cube roots of unity,

(1 - ω + ω^{2}) (1 - ω^{2} + ω^{4}) (1 - ω^{4} + ω^{8})

...to 2n factors

=

Join BYJU'S Learning Program

If 1,ω and ω2 are cube roots of unity, show that (1+ω)(1+ω2)(1+ω4)(1+ω8)=1.

Given,(1+w)(1+w2)(1+w4)(1+w8)considering first 2 terms, we get, (1+w)(1+w2)=1+w2+w+w3=0+w3=1similarly, considering second 2 terms, we get, (1+w4)(1+w8)=(1+w)(1+w2)=11×1=1Hence proved.

If $1, ω$ and $ω^{2}$ are cube roots of unity, show that $(1 + ω) (1 + ω^{2}) (1 + ω^{4}) (1 + ω^{8}) = 1$ .

Given,

$(1 + w) (1 + w^{2}) (1 + w^{4}) (1 + w^{8})$

considering first 2 terms, we get,

$(1 + w) (1 + w^{2}) = 1 + w^{2} + w + w^{3} = 0 + w^{3} = 1$

similarly, considering second 2 terms, we get,

$(1 + w^{4}) (1 + w^{8}) = (1 + w) (1 + w^{2}) = 1$

$1 \times 1 = 1$

Hence proved.