If x=a+b, y=α x+bβ and z=aβ+bα, where α and β are complex cube roots of unty,then
If x=a+by=aα+bβ and z=αβ+βα
Then xyz=(a+b)(aw+bw2)(aw2+bw),where α=w and β=w2 =(a+b)(a2+abw2+abw+b2)
This is a variable to variable question i.e. the question is invariables and the options
are in variables.So, we can assume and substitute any value for the variables.
Tricks: Put a=b=2
Then x=4,y=2(w+w2)=−2 and z=2(w2+w)=−2
∵ xyz=4(−2)(−2)=16 and (b) i.e. a3+b3=16.
This method works on elimination. (B) is the correct option because (A), (B) and (D) are
not equal to 16.