nope, the way u were trying doing it was the way i did.
note that (x+y)
3=x
3+3x
2y+3xy
2+y
3
this is the part u got wrong.
however if u look at drogonski's method,
he used the factorisation of cubes,
x
3+y
3=(x+y)(x
2+xy+y
2)
this shows two different ways of manipulating x
3+y
3 of which the latter was better for this question.