Question

if x+y+4=0 then find the value of x3+y3-12xy+64.

Answer 1

x+y = -4
(x+y)³ = (-4)³
x³ + y³ + 3xy(x+y) = -64
x³ + y³ + 3xy(-4) = -64
x³ + y³ - 12xy = -64
x³ + y³ - 12xy + 64 = 0

(a+b)³ = a³ + b³ + 3ab(a+b)

Answer 2

$x + y + 4 = 0 \\ ..x + y = - 4 \\ ..(x + y) {}^{3} = ( - 4 ) {}^{3} \\ .. {x}^{3} + {y}^{3} + 3xy(x + y) = - 64 \\ .. {x}^{3} + {y}^{3} + 3xy( - 4) = - 64 \\ .. {x}^{3} + {y}^{3} - 12xy = - 64 \\ .. {x}^{3} + {y}^{3} - 12xy + 64 = 0$
so the Value of you expression is 0......

hope this helps