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if x=a+b,y=aω+bω2,z=aω2+bω, then find the value of xyz ?
Dear Mahima,
As you know that 'w' is the cube root of unity:
so w3 = 1 & w2+w + 1 = 0;
so
(a+b)(aw+bw2)(aw2+bw)
= w2 (a+b)(a+bw)(aw+b)
=(a+b)(a+bw)(aw3+bw2)
=(a+b)(a+bw)(a+bw2) =a3+b3
Thus your required answer is : " a3+b3 "
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