Question 175687
1)  Factor by grouping:

x^3 + y^3 - x^2y - xy^2 =
(x^3 + y^3) + -(x^2y + xy^2).  The first is a sum of cubes:
[(x + y)(x^2 - xy + y^2)] + [-xy(x + y)].  They both have an x + y in common
(x + y)[(x^2 - xy + y^2) + (-xy)].  Now simplify:
(x + y ( x^2 - 2xy + y^2)

2)  This is a difference of squares:

81x^4 - 16y^4 =
(9x^2)^2 - (4y^2)^2 = (9x^2 - 4y^2)(9x^2 + 4y^2).  But the first is a difference of squares, so that can be factored again:

(3x - 2y)(3x + 2y)(9x^2 + 4y^2)

3)  Factor out a common factor, x:

x^3 + 2x^2 -255x =
x(x^2 + 2x - 255).  Can this be factored again?  255 = 15 * 17 which has a difference of two (the middle coefficient), so it can be factored:
x(x - 15)(x - 17)