Question

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

Answer 1

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

To Prove: $(a+b+c{)}^3=27abc$

Solution:

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

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

Cubing on both the sides,

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

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

or, $a+b+c+3\times a^{\frac{1}{3}}\times b^{\frac{1}{3}}\times (-c^{\frac{1}{3}})=0$ (from eqn.(1))

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

Cubing on both the sides,

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

or, $(a+b+c{)}^3=27abc$ proved.