Find the value of x^3+y^3-12xy+64 when x+y=-4

plz help!!!!!!

x^3 +y^3 - 12xy + 64
= x^3 + y^3 + 64 - 3(4xy) => Notice that the polynomial is in the form of:
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)
therefore it factors as:
x^3 + y^3 + 64 - 12xy
= (x + y + 4)(x^2 + y^2 + 16 - xy - 4y - 4x) => substitute for x + y = -4
= (-4 + 4)(x^2 + y^2 + 16 - xy - 4y - 4x)
= 0 * (x^2 + y^2 + 16 - xy - 4y - 4x)
= 0

Regards.

well... first off...

if x + y = -4 , isolate one variable...

(lets say u chose x)

the eqation now becomes x = -y + -4

(if u chose y.. the equation would be y = -x + -4 )

[i'll be using the x]

now that you hav isolated the x in the problem, move onto solving the bigger equation..

wherever u see the "x" .. plug in ( -y + -4 )

**dont forget the parantheses

it now becomes..

( -y + -4 )^3+y^3-12( -y + -4 )y+64

and solve for y...

:D

once u solve for y... plug THAT y-value in and solve for x!!

hope this helps!

(x+y)^3 ---> x^3 +3x^2y+3xy^2 +y^3 ---> x^3 +y^3 + 3xy(x+y)
using (x+Y) as -4, we get (x+y)^3 equals x^3 +Y^3 -12xy
so x^3 + y^3 -12xy +64 --->(x+y)^3 +64 ie -64+64 --->0

x^3+y^3-12xy+64
(x+y)^3-3xy(x+y)=formula
(x+y)^3-3xy(x+y)-12xy+64
(-4)^3-3xy(-4)-12xy+64
-64+12xy-12xy+64=0