Ejercicio 3:
x+y= -z
hallar k = x³+y³+z³/xyz
________
(x+y+z)³ = 0
(x+y+z)³ = x³+y³+z³+3(x+y)(x+z)(y+z)
0³= x³+y³+z³+3(-z)(-y)(-x)
x³+y³+z³ -3xyz = 0
x³+y³+z³ = 3xyz
Remplazando
K = 3xyz/xyz = 3
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Ejercicio 3:
x+y= -z
hallar k = x³+y³+z³/xyz
________
(x+y+z)³ = 0
(x+y+z)³ = x³+y³+z³+3(x+y)(x+z)(y+z)
0³= x³+y³+z³+3(-z)(-y)(-x)
x³+y³+z³ -3xyz = 0
x³+y³+z³ = 3xyz
Remplazando
K = 3xyz/xyz = 3
Respuesta = K = 3