## anonymous 5 years ago I need help trying to work this problem out, 8a^3+27? the result comes out to be (2a+3)(4a^2-6a+9). i just dont how you get that answer.

1. anonymous

Well you know that 8 and 27 are both cubes, (2, and 3 respectively) so you can start that way. Secondly you know that there aren't any middle terms, thus they must cancel in the cubing process. There is a formulaish way: $(\beta a + \gamma)(\beta^2a^2 - (\beta\gamma) a + \gamma^2)$ if the cube is $(\beta a^3 + \gamma^3)$ or $(\beta - \gamma)(\beta a^2 + (\beta\gamma)a + \gamma^2)$ if it is $(\beta^3 - \gamma^3)$

2. anonymous

thank you. this formula and cubing is a little confusing at times. But I am trying to plug it in to help show the work to get the result. so this is taking some time.