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+b^{3 "}

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