If 
where 
and 
then the value of 
is
We have,
= - 1
x2 + y2 = - xy
x2 + y2 + xy = 0
∴ Value of (x3 – y3) = (x – y) (x2 + y2 + xy)
= (x - y) × 0
= 0
Hence, option C is correct
33
If where and then the value of is
We have,
= - 1
x2 + y2 = - xy
x2 + y2 + xy = 0
∴ Value of (x3 – y3) = (x – y) (x2 + y2 + xy)
= (x - y) × 0
= 0
Hence, option C is correct