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I know about Pascal's triangle, but I don't know if that's what you're referring to.
Sure, do need an explanation of what it is, or do you have a problem you want me to look at?
Sure, do you know how to expand a binomial (something)^2?
Okay, let's start there then. Say you have something like (x + y)^2 and we want to expand that. You can rewrite (x + y)^2 so that it's like: (x + y)(x + y) Make sense so far?
Good, now we're going to multiply the first x in (x + y) to BOTH the (x + y) in the other term to get: x^2 + xy |dw:1358369073609:dw|
pascal's triangle is use to expand
Great, then we move on and do the same thing with the first y. and get xy + y^2, so in total we have x^2 + xy + xy + y^2
|dw:1358369198853:dw|do you see the pattern?
perfect, so we have x^2 + xy + xy + y^2 which equals x^2 + 2xy + y^2, which is what precal's picture shows
|dw:1358369387693:dw|then notice what is happening to the x powers and then the y powers
x powers start at 2 while y start at 0
this is the pattern you will follow for you problem
Now, if we wanted to take that further like (x +y)^3 (like your problem) we can do two things.
|dw:1358369548166:dw|you will use the 3rd row
We could multiply (x + y)(x^2 + 2xy + y^2) And do the EXACT same thing like how we were multiplying the x to the second term and then the y to the second term.
That would give us the correct answer, OR we could use the pattern precal is showing us.
be careful with your second term because of the negative, just simplify and you are done