Factorise.(x+y)^3 +(2y+3z)^3+ (3z+4x)^3 -3(x+y)(2y+3z)(3z+4x)
Note that this is just the same as
a^3+b^2+c^3-3abc = (a+b+c)(a^2-ab-ac+b^2-bc+c^2)
which you get using the sum-of-cubes factorization and some ingenuity.
These look more like contest problems than homework assignments!