Question

If $a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}}=0$ then $(a+b+c{)}^3=27abc$ how?

Answer 1

Let $\begin{array}{cccccc}{a}^{y_3}& +& b^{y_3}& +& c^{y_3}& =0\\ \downarrow & & \downarrow & & \downarrow & \\ A& & B& & C& \end{array}$
Then we know $A^3+B^3+C^3=3ABC$ when $(A+B+C=0)$
That is, $(a^{y_3}{)}^3+(b^{y_3}{)}^3+(c^{y_3}{)}^3=3a^{y_3}{b}^{y_3}{c}^{y_3}$
$\Rightarrow a+b+c=3a^{y_3}{b}^{y_3}{c}^{y_3}$
Cubing both sides,
$\Rightarrow (a+b+c{)}^3=27abc$ Proved.