Thread: Help! How to foil this cube expression??

Consider:

(x+y)^2

= (x+y)(x+y)

= x^2 + xy + xy + y^2

= x^2 +2xy + y^2

Question:

(-2-5)^-3

= 1/(-2-5)^3

= 1/(-2-5)(-2-5)(-2-5)

= 1/?

How to foil such an expression?

(x+y)^3

= (x+y)(x+y)(x+y)

= ?

As again, thanks for the help.

Originally Posted by shenton

Consider:

(x+y)^2

= (x+y)(x+y)

= x^2 + xy + xy + y^2

= x^2 +2xy + y^2

Question:

(-2-5)^-3

= 1/(-2-5)^3

= 1/(-2-5)(-2-5)(-2-5)

= 1/?

How to foil such an expression?

(x+y)^3

= (x+y)(x+y)(x+y)

= ?

As again, thanks for the help.

You can either look it up as a formula or do it in steps. As you said:

(x + y)^2 = x^2 + 2xy + y^2. Thus:

(x + y)^3 = (x + y)(x^2 + 2xy + y^2) = x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2)

= x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3

So
(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

Similarly:
(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

-Dan

think of the distributive property:

x(a+b)=ax+bx

now try: (a+b)(x+y)

use distributive property: x(a+b)+y(a+b)

use distributive property again: ax+bx+ay+by


now try: (x+y)^3

rewrite: (x+y)(x+y)(x+y)

use foil: (x^2+2xy+y^2)(x+y)

use distributive property: x^2(x+y)+2xy(x+y)+y^2(x+y)

can you solve from there?

With a little work you can find the following:
(x+y)^2 = x^2 + 2xy + y^2

(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5

etc.

Note that in the expansion for (x+y)^n each term takes the form cx^a*y^b with a+b = n, and all possible terms are listed.

As far as the constants are concerned there are various ways to produce them, the simplest being the combinitorial function, usually written as

nCr = (n!)/[(r!)(n - r)!]

You can also use Pascal's Triangle. (Which, if I may be allowed to brag, I discovered on my own. )

So, to write out, say, (x+y)^4 we do the following:

(x+y)^4 = (4C0)*x^4*y^0 + (4C1)x^(4-1)*y^1 + (4C2)x^(4-2)*y^2 + (4C3)x^(4-3)*y^3 + (4C4)x^(4-4)*y^4

(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

-Dan

Originally Posted by topsquark

(x + y)^2 = x^2 + 2xy + y^2. Thus:

(x + y)^3 = (x + y)(x^2 + 2xy + y^2) = x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2)

= x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3

So
(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

Similarly:
(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

Thanks for the step by step proof. That is very helpful in understanding.

Originally Posted by Quick

now try: (x+y)^3

rewrite: (x+y)(x+y)(x+y)

use foil: (x^2+2xy+y^2)(x+y)

use distributive property: x^2(x+y)+2xy(x+y)+y^2(x+y)

Thanks again. I think I can solve from there.