Alg3.3.8

If $a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}}=0\,$ Show that $(a+b+c)^3=27abc\,$

Given $a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}}=0\,$

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

Cubing on both sides,

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

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

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

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

Again cubing on both sides,we get

$(a+b+c)^3=-27(ab(-c))=27abc\,$