(i) Given
x + y + 4 = 0
∴ x3+y3+(4)3=3xy(4) …….(1)
[Using the identity, if a+b+c=0, then a3+b3+c3=3abc]
=12xy
Now,
x3+y3−12xy+64=x3+y3+64−12xy
=12xy−12xy=0 [from Eq. (i) ]
(ii) Given,
x−2y−6=0
∴ x3+(−2y)3+(−6)3=3x(−2y)(−6)
[Using the identity, if a+b+c=0, then a3+b3+c3=3abc]
x3−8y3−216=36xy …..(i)
Now, x3−8y3−36xy−216
=x3−8y3−216−36xy
=36xy−36xy=0 [from Eq. (i)]