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

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

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

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

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

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

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

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

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

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

ParthKohliBest 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
 one year ago

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

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

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

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

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

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

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

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

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

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

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