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I know this simplifies to\[\left(\dfrac{2}{3}x + \dfrac{5}{3}y\right)^3 + \rm blah\]
So for the first two terms, should I use \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)?

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That'd make stuff very long.
its like \((a+b)^3+(c-b)^3-(a+c)^3\)
Yeah. :-|
Wait, so\[(a + c)(a^2 + 2ab + b^2 - (a+b)(c -b) + c^2 - 2cb + b^2) - (a + c)^3\]
\(\large = 3(a+b)(a+c)(b-c)\)
How did you get that?
It's correct, but is it some sort of identity taught?
ok, first take last 2 terms, (c-b)^3-(a+c)^3 = ... ? use A^3-B^3 =...
not identity, you need to work that out...
its quite long though
\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]So here,\[= -(b +a)(\cdots)\]Err
No shortcuts?
none that i can think of.
There's always a shortcut in exam questions. It's not like an Olympiad one.
then you must be having choices ?
No choices given. That's all.
:-(
then i'll wait for someone to suggest a shortcut.....
OK, thanks for the support :-)
Is Pascal's Triangle any useful? http://www.mathsisfun.com/pascals-triangle.html
I usually use that, but i don't know if its going to help you?
@hartnn Oh BTW, I found the solution.
any shortcuts ?
http://math.stackexchange.com/questions/361002/factoring-a-long-expression-in-the-form-ab3-c-b3-cb3?noredirect=1#comment775818_361002 Yeah, sort of.
isn't that exactly what i told you ?
:-P

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