If w is a complex cube root of unity then show that (2-w)(2-w^2)(2-w^10)(2-w^11)=49(2w)(2w2)(2w10)(2w11)=49?

2 Answers
Aug 7, 2018

Kindly refer to the Explanation.

Explanation:

Since, w in CC, w=root(3)(1)," we have, "w^3-1=0.

:. (w-1)(w^2+w+1)=0, or, w^2+w=-1......(star).

"The Expression"={(2-w)(2-w^2)}{(2-w^10)(2-w^11)},

={4-2(w+w^2)+w^3}{4-2(w^10+w^11)+w^21},

={4-2(-1)+1}{4-2w^9(w+w^2)+1^7},

=(7){4-2(1^3)(-1)+1},

=(7)(7),

=49, as desred!

Aug 8, 2018

omega being imaginary cube root of unity,we know
omega^3=1and1+omega+omega^2=0

"Now the given expression"

=(2-w)(2-w^2)(2-w^10)(2-w^11)

=(2-w)(2-w^2)(2-(w^3)^3*w)(2-(w^3)^3*w^2)

=(2-w)(2-w^2)(2-w)(2-w^2)

=(4-2w-2w^2+w^3)^2

=(4-2(w+w^2)+1)^2

=(4-2((-1))+1)^2

=7^2=49