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

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Answer 1 · 2018-08-07T18:28:39

Kindly refer to the 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!

Answer 2 · 2018-08-08T14:30:52

#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#