Question #bc3b5

1 Answer
Jul 31, 2017

#256y^8+2048xy^7+7168x^2y^6+14336x^3y^5+17920x^4y^4+14335x^5y^3+7168x^6y^2+2048x^7y+256x^8#

Explanation:

#((8), (0)) = 1#
#((8), (1)) = 8#
#((8), (2)) = 28#
#((8), (3)) = 56#
#((8), (4)) = 70#
#((8), (5)) = 56#
#((8), (6)) = 28#
#((8), (7)) = 8#
#((8), (8)) = 1#

Binomial expansion:
http://www.purplemath.com/modules/binomial.htm

So, here we have:
#(-2x-2y)^8=1(-2x)^0(-2y)^8+8(-2x)^1(-2y)^7+28(-2x)^2(-2y)^6+56(-2x)^3(-2y)^5+70(-2x)^4(-2y)^4+56(-2x)^5(-2y)^3+28(-2x)^6(-2y)^2+8(-2x)^7(-2y)^1+1(-2x)^8(-2y)^0#

It seems like a lot, but can easily be simplified down into: #256y^8+2048xy^7+7168x^2y^6+14336x^3y^5+17920x^4y^4+14335x^5y^3+7168x^6y^2+2048x^7y+256x^8#