If $$x^{1/3} + y^{1/3} - z^{1/3}$$​ = 0, then find the value of (x+y-z)³ + 27 xyz.

Correct option is B

Given:

​$$x^{1/3} + y^{1/3} - z^{1/3} = 0$$​​

Solution:

Rearranging the equation, we get:

​$$x^{1/3} + y^{1/3} = z^{1/3}$$​​

Cubing both sides, we get:

​$$(x^{1/3} + y^{1/3})^3 = (z^{1/3})^3$$​​

​$$x + y + 3(x^{1/3})(y^{1/3})(x^{1/3} + y^{1/3}) = z$$​​

Substituting $$x^{1/3} + y^{1/3} = z^{1/3}$$​, we simplify to:

​$$x + y + 3(xyz)^{1/3} = z$$​​

Rearranging the equation:

​$$x + y - z = -3(xyz)^{1/3}$$​​

Now, cubing both sides, we get:

​$$(x + y - z)^3 = (-3(xyz)^{1/3})^3\\\ \\(x + y - z)^3 = -27xyz\\\ \\Thus, (x + y - z)^3 + 27xyz = -27xyz + 27xyz = 0 .$$​​

Hence, the value of $$(x + y - z)^3 + 27xyz = 0$$​.