If $$x^{\frac{1}{3}} + y^{\frac{1}{3}} = z^{\frac{1}{3}}$$, then $$(x + y - z)^3 + 27 xyz$$ is equal to

Given,       $$x^{\ \frac{\ 1}{3}}+\ y^{\ \frac{\ 1}{3}} =z^{\ \frac{\ 1}{3}}$$

         $$\left(x^{\ \frac{\ 1}{3}}+\ y^{\ \frac{\ 1}{3}}\right)^3=z^{\ \frac{\ 3}{3}}$$

$$x^{\ \frac{\ 3}{3}}+y^{\ \frac{\ 3}{3}}+3.x^{\ \frac{\ 1}{3}}.y^{\ \frac{\ 1}{3}}\left(x^{\ \frac{\ 1}{3}}+y^{\ \frac{\ 1}{3}}\right)=z^{\ }$$

$$x+y+3.x^{\ \frac{\ 1}{3}}.y^{\ \frac{\ 1}{3}}\left(z^{\ \frac{\ 1}{3}}\right)=z^{\ }$$

                   $$=$$>     $$x+y-z= -3.x^{\ \frac{\ 1}{3}}.y^{\ \frac{\ 1}{3}}.z^{\ \frac{\ 1}{3}}$$

Therefore    $$\left(x+y-z\right)^3+27xyz\ =\ \left(-3.x^{\ \frac{\ 1}{3}}.y^{\ \frac{\ 1}{3}}.z^{\ \frac{\ 1}{3}}\right)^3 \ +27xyz$$  =  $$-27xyz+27xyz$$ =  0