identity

[solakis]

Given:
$a+b+c-3(abc)^\dfrac{1}{3}=0$ and $\neg((a)^\dfrac{1}{3}=(b)^\dfrac{1}{3}) $ and $\neg((b)^\dfrac{1}{3}=(c)^\dfrac{1}{3})$ and
$\neg((c)^\dfrac{1}{3}=(a)^\dfrac{1}{3})$
Then prove:

$(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$

 

[solakis]

Given:
$a+b+c-3(abc)^\dfrac{1}{3}=0$...............................................................(1)

$\neg((a)^\dfrac{1}{3}=(b)^\dfrac{1}{3}) $ ........................................(2)
$\neg((b)^\dfrac{1}{3}=(c)^\dfrac{1}{3})$ ........................................(3)
$\neg((c)^\dfrac{1}{3}=(a)^\dfrac{1}{3})$............................................(4)
Then prove:

$(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$

Spoiler

Let : $x=a^\dfrac{1}{3},y=b^\dfrac{1}{3},z=c^\dfrac{1}{3}$......................................(5)
Use the following identity:

$(x^3+y^3+z^3-3xyz)=\dfrac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$.................................(6)

Use (1),(2),(3) ,(4),(5) and (6) becomes:

$ (a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$
Since:
$x+y+z=0$