Question

if x^1/3+y^1/3+z^1/3=0 then find the value of (x+y+z)^3

Answer 1

x⅓ + y⅓ + z⅓ = 0
Let x⅓ = a
y⅓ = b
z⅓ = c
then , a + b + c = 0
we know, when, a + b + c = 0
then, a³ + b³ + c³ = 3abc
hence,
(x⅓)³ + (y⅓)³ + (z⅓)³ = 3(x⅓)(y⅓)(z⅓)

x + y + z = 3(xyz)⅓

take cube both sides,

(x + y + z)³ = (3)³{ (xyz)⅓}³

(x + y + z)³ = 27xyz(answer )

Answer 2

Answer:

x⅓ + y⅓ + z⅓ = 0

Let x⅓ = a

y⅓ = b

z⅓ = c

then , a + b + c = 0

we know, when, a + b + c = 0

then, a³ + b³ + c³ = 3abc

hence,

(x⅓)³ + (y⅓)³ + (z⅓)³ = 3(x⅓)(y⅓)(z⅓)

x + y + z = 3(xyz)⅓

take cube both sides,

(x + y + z)³ = (3)³{ (xyz)⅓}³

(x + y + z)³ = 27xyz(answer )