Question

Factor.

Answer 1

I know this simplifies to\[\left(\dfrac{2}{3}x + \dfrac{5}{3}y\right)^3 + \rm blah\]

Answer 2

So for the first two terms, should I use \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)?

Answer 3

That'd make stuff very long.

Answer 4

its like
\((a+b)^3+(c-b)^3-(a+c)^3\)

Answer 5

Yeah. :-|

Answer 6

Wait, so\[(a + c)(a^2 + 2ab + b^2 - (a+b)(c -b) + c^2 - 2cb + b^2) - (a + c)^3\]

Answer 7

\(\large = 3(a+b)(a+c)(b-c)\)

Answer 8

How did you get that?

Answer 9

It's correct, but is it some sort of identity taught?

Answer 10

ok, first take last 2 terms,
(c-b)^3-(a+c)^3 = ... ?
use A^3-B^3 =...

Answer 11

not identity, you need to work that out...

Answer 12

its quite long though

Answer 13

\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]So here,\[= -(b +a)(\cdots)\]Err

Answer 14

No shortcuts?

Answer 15

none that i can think of.

Answer 16

There's always a shortcut in exam questions. It's not like an Olympiad one.

Answer 17

then you must be having choices ?

Answer 18

No choices given. That's all.

Answer 19

:-(

Answer 20

then i'll wait for someone to suggest a shortcut.....

Answer 21

OK, thanks for the support :-)

Answer 22

Is Pascal's Triangle any useful?
http://www.mathsisfun.com/pascals-triangle.html

Answer 23

I usually use that, but i don't know if its going to help you?

Answer 24

@hartnn Oh BTW, I found the solution.

Answer 25

any shortcuts ?

Answer 26

http://math.stackexchange.com/questions/361002/factoring-a-long-expression-in-the-form-ab3-c-b3-cb3?noredirect=1#comment775818_361002

Yeah, sort of.

Answer 27

isn't that exactly what i told you ?

Answer 28

:-P