If $a^{1/3}+b^{1/3}+c^{1/3}=0$ , then ${\left(a+b+c\right)}^3$

$a^{\frac{1}{3}}+b^{\frac{1}{3}}=-c^{\frac{1}{3}}$

Cubing on both sides

${\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)}^3={\left(-c^{\frac{1}{3}}\right)}^3$

${\left(a^{\frac{1}{3}}\right)}^3+{\left(b^{\frac{1}{3}}\right)}^3+3.a^{\frac{1}{3}}b{.}^{\frac{1}{3}}\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)=-c$

$a+b+3.a^{\frac{1}{3}}.b^{\frac{1}{3}}\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)=-c$

$a+b+c=3a^{\frac{1}{3}}{b}^{\frac{1}{3}}{c}^{\frac{1}{3}}$

Again cubing on both sides

${\left(a+b+c\right)}^3=27abc$