If 1- - -2 are the cube roots of unity- then prove that 12 - - - 11 - 2- -1-2

Fact: 1+ω+ω^2=0 So LHS =12 + ω + 11 + 2ω =1+2ω+2+ω(2+ω)(1+2ω) =3(1+ω)2+5ω+2ω^2=3(1+ω)5ω+2(1+ω^2) =3(1+ω)5ω 2ω=3(1+ω)3ω=1+ωω= ω^2ω= ω=1+ω ...

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Byju's Answer

Standard XII

Mathematics

Cube Root of a Complex Number

If 1, ω, ω2...

Question

If $1, ω, ω^{2}$ are the cube roots of unity, then prove that $\frac{1}{2 + ω} + \frac{1}{1 + 2 ω} = 1 + ω^{2}$

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Solution

Fact: $1 + ω + ω^{2} = 0$
So LHS
$= \frac{1}{2 + ω} + \frac{1}{1 + 2 ω}$

$= \frac{1 + 2 ω + 2 + ω}{(2 + ω) (1 + 2 ω)}$

$= \frac{3 (1 + ω)}{2 + 5 ω + 2 ω^{2}} = \frac{3 (1 + ω)}{5 ω + 2 (1 + ω^{2})}$

$= \frac{3 (1 + ω)}{5 ω - 2 ω} = \frac{3 (1 + ω)}{3 ω} = \frac{1 + ω}{ω} = \frac{- ω^{2}}{ω} = - ω = 1 + ω^{2}$

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If 1,ω,ω2 are the cube roots of unity, then prove that 12+ω+11+2ω=1+ω2

Fact: 1+ω+ω2=0So LHS=12+ω+11+2ω=1+2ω+2+ω(2+ω)(1+2ω)=3(1+ω)2+5ω+2ω2=3(1+ω)5ω+2(1+ω2)=3(1+ω)5ω−2ω=3(1+ω)3ω=1+ωω=−ω2ω=−ω=1+ω2

If $1, ω, ω^{2}$ are the cube roots of unity, then prove that $\frac{1}{2 + ω} + \frac{1}{1 + 2 ω} = 1 + ω^{2}$