Proof $\sqrt[3]{a-b} + \sqrt[3]{b-c} + \sqrt[3]{c-a} \neq 0$

Question

Let $a,b,c \in R$, and they are not equal.

Proof $\sqrt[3]{a-b} + \sqrt[3]{b-c} + \sqrt[3]{c-a} \neq 0$

I can prove this statement by contradiction ($\sqrt[3]{a-b} + \sqrt[3]{b-c} + \sqrt[3]{c-a} = 0 \Rightarrow \sqrt[3]{a-b} + \sqrt[3]{b-c} = - \sqrt[3]{c-a} \Rightarrow 3(\sqrt[3]{(a-b)^{2}(b-c)} + \sqrt[3]{(a-b)(b-c)^{2}})=0 \Rightarrow \dots)$

But I want a solution with Euler's formula ($x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-xz-yz)$). If we set $x=\sqrt[3]{a-b}, y=\sqrt[3]{b-c}, z= \sqrt[3]{c-a}$, then $$0-3\sqrt[3]{(a-b)(b-c)(a-c)} = (\sqrt[3]{a-b} + \sqrt[3]{b-c} + \sqrt[3]{c-a})(A)$$

Since $3\sqrt[3]{(a-b)(b-c)(a-c)}\neq 0$, if I show $A\neq 0$, the proof is complete, but I can't proov it. Can you help me?

Answer 1

The constraint for $a$, $b$ and $c$ being not all equal are not strong enough.

\begin{align} x^3+y^3+z^3-3xyz &= (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \\ &= (x+y+z) \times \frac{(x-y)^2+(y-z)^2+(z-x)^2}{2} \\ x &= \sqrt[3]{a-b} \\ y &= \sqrt[3]{b-c} \\ z &= \sqrt[3]{c-a} \\ x^3+y^3+z^3 &= 0 \\ \end{align}

Note that

• $xyz = 0 \implies a=b \lor b=c \lor c=a$

• $x=y=z \implies a=b=c$

Distinct $a,b,c$ guarantee $$xyz \ne 0$$ and $$\frac{(x-y)^2+(y-z)^2+(z-x)^2}{2} \ne 0$$

Answer 2

Say there exist $a,b,c$ not equal such that

$$\sqrt[3]{a-b} + \sqrt[3]{b-c} + \sqrt[3]{c-a} = 0$$

Then we have

$$\sqrt[3]{a-b} + \sqrt[3]{b-c} = \sqrt[3]{a-c}$$

Write $x= \sqrt[3]{a-b}$ and $y= \sqrt[3]{b-c}$, then we have $$(x+y)^3 = x^3+y^3$$

So $3xy(x+y)=0$ which means that $a=b$ or $b= c$ or $c=a$. A contradiction.

Answer 3

i would substitute $$x=\sqrt[3]{a-b},y=\sqrt[3]{b-c},z=\sqrt[3]{c-a}$$ then we get $$x^3+y^3+z^3=...$$?