asked by Jessy
  1. You have the usual sum/difference of squares/cubes.

    (xz^2)^3 - (3y^3)^3
    = (xz^2-3y^3)((xz^2)^2 + (xz^2)(3y^3) + (3y^3)^2)

    Similarly, the others are

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

    Just apply your formulas. If you get stuck, come on back and say where.

    posted by Steve

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