Question

If $a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}}=0$ then $\left(a+b+c\right)=27abc$. Prove $x$ by using appropriate formula .

Answer 1

if $x+y+z=0$, then $x^3+y^3+z^3$ = $3xyz$

in our case,

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

$\therefore (a^{1/3}{)}^3+(b^{1/3}{)}^3+(c^{1/3}{)}^3=3a^{1/3}{b}^{1/3}{c}^{1/3}$

or $(a+b+c)=3(abc{)}^{1/3}$

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

proved.