Question

If $a^{1/3}+b^{1/3}+c^{1/3}=0$,then show that $(a+b+c{)}^3=27abc$

Answer 1

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

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

Cubing both sides, we get

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

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

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

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

Cubing both sides, we get

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

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

Hence proved.