(x+y)^3= ? or (x+y)³= ?

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

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

Simplify?:
= (x + y)^3
= (x + y)(x + y)(x + y)
= (x^2 + 2xy + y^2)(x + y)
= x^3 + 3x^2y + 3xy^2 + y^3

(x + y) (x² + 2xy + y²)

x³ + 2x²y + xy²
____x²y + 2xy² + y³

x³ + 3x²y + 3xy³ + y³

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

(x+y)³ = (x+y) (x+y) (x+y)
it really is just whatever is within the parenthesis times itself 3 times
(x+y)³ =(x^2+2xy+y^2)(x+y)
foil the first 2
(x+y)³ =x^3+2x^2y+xy^2+x^2y+2xy^2+y^3
then foil 3rd one
(x+y)³ = x^3+3x^2y+3xy^2+y^3
then you put like terms together

(x^2+2xy+y^2)(x+Y)

x^3+2x^2y+y^2x+x^2y+2y^2x+y^3

stop cheating ;)

Start by writing the x terms in descreasing powers and the y terms in increasing powers, so they each add up to 3:

x^3 + x²y + xy² + y^3

Now take the values from the 3rd row of Pascal's triangle (1 3 3 1) and make them the coefficients:

x^3 + 3x²y + 3xy² + y^3