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Mathematics 20 Online
OpenStudy (anonymous):

Expand (x+y)^3?

OpenStudy (aroub):

x^3+3a^2b+3ab^2+b^3

OpenStudy (anonymous):

Urge to delete rising..

OpenStudy (anonymous):

To expand you basically do a lot of distribution

myininaya (myininaya):

1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 this some of the levels of the pascal triangle this an easy way to memorize coefficients of (x+y)^n

OpenStudy (anonymous):

Memorize schemorize.. Nobody wants to keep this in their brain

OpenStudy (anonymous):

Ha no I don't do well with memorization

OpenStudy (anonymous):

You don't need to memorize pascal, just add two term from above

myininaya (myininaya):

\[(x+y)^n=x^n++x^{n-1}y+x^{n-2}y^2+ \cdot \cdot \cdot +x^2y^{n-2}+xy^{n-1}+y^n\]

myininaya (myininaya):

no seriously it is the most easy thing to memorize

myininaya (myininaya):

do you not see an easy pattern all you are doing is adding the two number above to find the one below

OpenStudy (anonymous):

\[(x+y)^3\]\[=(x+y)(x+y)(x+y)\]\[=(x^2 + 2xy + y^2)(x+y)\]\[=x^2(x+y) + 2xy(x+y) + y^2(x+y)\]\[=x^3 + x^2y + 2x^2y + 2xy^2 + xy^2 + y^3\]\[=x^3+ 3x^2y + 3xy^2 + y^3\]

OpenStudy (anonymous):

You already know from your last problem how to do the first two factors (x+y)(x+y)

OpenStudy (anonymous):

Then you take that result and distribute the last factor (x+y) to each term in that result

myininaya (myininaya):

1 1 2 1 1 (1+2) (2+1) 1 1 1+(1+2) (2+1+1+2) 1+(2+1 1 so on.....................

OpenStudy (anonymous):

Then you distribute and combine like terms

OpenStudy (anonymous):

Okay I think I get it

OpenStudy (anonymous):

Make sure, or figure out what you're shaky on

OpenStudy (anonymous):

I worked through it by myself and I understand it for sure. Thanks!

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