Grade 12Algebra
If 1, w, w2 are cube roots of unity, prove the following:(1 - w + w^2) (1+ w - w^2) = 4
If 1, w, w2 are cube roots of unity, prove the following:(1 - w + w^2) (1+ w - w^2) = 4
3 Answers
Harshit Singh
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
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
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
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
Please aprove
Thanks
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