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

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The discussion focuses on proving that $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$ under the condition that $a+b+c-3(abc)^\dfrac{1}{3}=0$ and the cube roots of a, b, and c are distinct. By letting $x=a^\dfrac{1}{3}, y=b^\dfrac{1}{3}, z=c^\dfrac{1}{3}$, the identity relating the sum of cubes to their product is applied. Given the established conditions, it follows that $x+y+z=0$, leading to the conclusion that the sum of the cube roots equals zero. This proof effectively demonstrates the relationship between the variables under the specified constraints.
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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:

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

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