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# If 1, w, w2 are cube roots of unity, prove the following:(1 - w + w^2) (1+ w - w^2) = 4

Harshit Singh
askIITians Faculty 5962 Points
11 months ago
Dear Student

1+ w + w^2 = 0
so 1 + w = – w^2
1+ w^2 = – w
(1 - w + w^2)= –2w
(1+ w - w^2)= –2w^2
(1 - w + w^2) (1+ w - w^2)= (-2w)( –2w^2)
= 4w^3
and we know that w^3=1
Hence, proved

Thanks
Anshu Kumar
14 Points
9 months ago
As we know from cube root of unity is
the sum of cube root of unity is zero and product of cube root of unity is one, so
1+w+w^2 = 0 & w^3 = 1  which implies that
1+w = -w^2  …....(i) and
1+w^2 = -w ….....(ii)
Now form Left Hand side of given equation we have
(1-w+w^2)(1+w-w^2) we can write it as follows
(1+w^2-w)(1+w-w^2)
(-w-w)(-w^2-w^2) …........ from (i) & (ii)
(-2w)(-2w^2) = 4w^3 =4*1 = 4 .
Hence we get Right Hand Side of given Equation.

Ram Kushwah
110 Points
9 months ago
We know that
If 1,ω,ω² are roots of unity then
ω³=1..................................(1)
1+ω+ω²=0........................(2)
Now
(1-ω+ω²)( 1+ω-ω²)
= ( 1+ω+ω²-2ω)(1+ω+ω²-2ω²)

putting 1+ω+ω²=0 we get
=(0-2ω)(0-2ω²)
=(-2 x -2) ω³
=4ω³
=4  (  As ω ³=1)
Hence proved