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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 If 1, w, w2 are cube roots of unity, prove the following:(1 - w + w^2) (1+ w - w^2) = 4
Dear Student1+ w + w^2 = 0so 1 + w = – w^21+ 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^3and we know that w^3=1Hence, provedThanks
As we know from cube root of unity isthe sum of cube root of unity is zero and product of cube root of unity is one, so1+w+w^2 = 0 & w^3 = 1 which implies that1+w = -w^2 …....(i) and1+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.
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 aproveThanks
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