Question

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

Answer 1

x^⅓+ y^⅓+ z^⅓ = 0

(x^⅓)³+(y^⅓)³+(z^⅓)³= 3×{x^⅓}×{y^⅓}×{z^⅓}

[using the formula, if (a+b+c)=0
then a³+b³+c³=3abc]
= x+y+z = 3 ³√xyz
taking cube root of both sides we get ,
=(x+y+z)³=(3 ³√xyz)³
=(x+y+z)³=27xyz
Hope THIS HELPS YOU

Answer 2

x^1/3 + y^1/3 + z^1/3 = 0
let :
x^1/3 = a ,
y^1/3 = b,
z^1/3 = c

as , a + b + c = 0
then, a + b = -c
taking the cube both sides,
(a+b)^3 = (-c)^3

[identity = (a+b)^3 = a^3 + b^3 + 3a^2b+ 3ab^2 ]
a^3 + b^3 + 3a^2b + 3ab^2 = -c^3
a^3 + b^3 + c^3 = -3a^2 + 3ab^2
= -3ab(b+a)
= -3ab(-c)
= 3abc
(a^3 + b^3 + c^3) = (3abc)
taking cube both sides
(a^3 + b^3 + c^3)^3 = (3abc)^3
=27a^3 b^3 c^3
put the value of a,b and c
(x^1/3*3 + y^1/3*3 + z^1/3*3)
(powers divided)
therefore, =27(x*y*z)

=27xyz