Proving $\sqrt[3]{a-b}+\sqrt[3]{b-c}+\sqrt[3]{c-a}\ne 0$ with 3 Real Numbers

[anemone]

Let $a,\,b$ and $c$ be three distinct real numbers.

Prove that $\sqrt[3]{a-b}+\sqrt[3]{b-c}+\sqrt[3]{c-a}\ne 0$.

 

[MarkFL]

My solution:

Spoiler

Let's write the inequality as:

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

Cube both sides:

$$a-b+3\sqrt[3]{(a-b)^2}\sqrt[3]{b-c}+3\sqrt[3]{a-b}\sqrt[3]{(b-c)^2}+b-c\ne a-c$$

Combine like terms and then divide through by 3:

$$\sqrt[3]{(a-b)^2}\sqrt[3]{b-c}+\sqrt[3]{a-b}\sqrt[3]{(b-c)^2}\ne0$$

Divide through by $$\sqrt[3]{a-b}\sqrt[3]{b-c}\ne0$$ (recall the 3 variables are distinct)

$$\sqrt[3]{a-b}+\sqrt[3]{b-c}\ne0$$

Rearrange:

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

Cube:

$$a-b\ne c-b$$

Add through by $b$:

$$a\ne c$$

This is true, hence the original inequality is true. :)

 

[kaliprasad]

Let $a,\,b$ and $c$ be three distinct real numbers.

Prove that $\sqrt[3]{a-b}+\sqrt[3]{b-c}+\sqrt[3]{c-a}\ne 0$.

Spoiler

using if $x+y+z=0$ => $x^3+y^3+z^3= 3xyz$
we get
$\sqrt[3]{a-b}+\sqrt[3]{b-c}+\sqrt[3]{c-a}= 0$
=> $(a-b)+ (b-c)+(c-a) = \sqrt[3]{(a-b)(b-c)(c-a)}$
or $\sqrt[3]{(a-b)(b-c)(c-a)} = 0$
or $(a-b)(b-c)(c-a) = 0$ => $a\,b\,c$ cannot be distinct
as they are distinct given expression cannot be zero