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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I know this simplifies to\[\left(\dfrac{2}{3}x + \dfrac{5}{3}y\right)^3 + \rm blah\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0So for the first two terms, should I use \(a^3 + b^3 = (a + b)(a^2  ab + b^2)\)?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0That'd make stuff very long.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2its like \((a+b)^3+(cb)^3(a+c)^3\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Wait, so\[(a + c)(a^2 + 2ab + b^2  (a+b)(c b) + c^2  2cb + b^2)  (a + c)^3\]

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2\(\large = 3(a+b)(a+c)(bc)\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0How did you get that?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0It's correct, but is it some sort of identity taught?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2ok, first take last 2 terms, (cb)^3(a+c)^3 = ... ? use A^3B^3 =...

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2not identity, you need to work that out...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[a^3  b^3 = (a  b)(a^2 + ab + b^2)\]So here,\[= (b +a)(\cdots)\]Err

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2none that i can think of.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0There's always a shortcut in exam questions. It's not like an Olympiad one.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2then you must be having choices ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0No choices given. That's all.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2then i'll wait for someone to suggest a shortcut.....

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0OK, thanks for the support :)

some_someone
 one year ago
Best ResponseYou've already chosen the best response.0Is Pascal's Triangle any useful? http://www.mathsisfun.com/pascalstriangle.html

some_someone
 one year ago
Best ResponseYou've already chosen the best response.0I usually use that, but i don't know if its going to help you?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0@hartnn Oh BTW, I found the solution.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0http://math.stackexchange.com/questions/361002/factoringalongexpressionintheformab3cb3cb3?noredirect=1#comment775818_361002 Yeah, sort of.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2isn't that exactly what i told you ?
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