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

$$x^{\frac{1}{3}} + y^{\frac{1}{3}} - z^{\frac{1}{3}} = 0$$

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

Cubing both sides

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

$$(x^{\frac{1}{3}})^3 + (y^{\frac{1}{3}})^3+3(x^{\frac{1}{3}})^2y^{\frac{1}{3}}+3(y^{\frac{1}{3}})^2x^{\frac{1}{3}} =(z^{\frac{1}{3}})^3$$

$$x + y+3(x^{\frac{2}{3}})y^{\frac{1}{3}}+3(y^{\frac{2}{3}})x^{\frac{1}{3}} =z$$

$$x + y-z+3(x^{\frac{2}{3}})y^{\frac{1}{3}}+3(y^{\frac{2}{3}})x^{\frac{1}{3}} =0$$

$$x + y-z+3(x^\frac13)(y^\frac13)(x^\frac13+y^\frac13) =0$$

$$x + y-z+3(x^\frac13)(y^\frac13)(z^\frac13) =0$$

$$x + y-z = - 3(x^\frac13)(y^\frac13)(z^\frac13)$$

cubing both sides

$$(x + y - z)^3 + 27 xyz=0$$.