Question

Factor (3x+2y)^3+(y-3)^3

Answer 1

sum of cubes:
a^3 +b^3 = (a+b)(a^2-ab+b^2)

Answer 2

I know the formula but i end up getting
(3x+3y-3)(9x+4y-3xy-9x+2y^2-6y+y^2-9), and then i simply the first parenthesis which gives me 3(x+y-1) but according to the book it should be 9(x+y-1), and i have no idea how they got that, and i know the second part is wrong too

Answer 3

the 9 comes from factoring out another 3 from 2nd part

double check the 2nd part (a^2 -ab +b^2)
(3x+2y)^2 = 9x^2 +12xy +4y^2
(3x+2y)(y-3) = 3xy-9x-6y+2y^2
(y-3)^2 = y^2 -6y +9

Answer 4

Well i'm not sure still where the 9 comes from can you show me that........and when i combined like terms for the second part i got
(9x^2+9xy+7y^2-12y-9x+9)............is that right, and now i see that i can factor out the 9 in those so that will give me
(x^2+xy+7y^2-12y-x+1)........but that's not the answer in the book tho, so i'm not sure where i messed up