# Thread: If a^3+b^3+c^3=3abc, then prove that a=b=c.

1. ## If a^3+b^3+c^3=3abc, then prove that a=b=c.

If $a^3+b^3+c^3=3abc$, then prove that $a=b=c$.

$\displaystyle \text{By the identity, }a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca)\text{ we find either }a+b+c=0\text{ or }a^2+b^2+c^2+ab+bc+ca=0.\text{ the first thing is false since we arrive at a contradiction.}$
\displaystyle \text{ I tried many things even by expressing }\begin{aligned} a^2+b^2+c^2+ab+bc+ca&=a^2+b^2+c^2+2ab+2bc+ca-ab-bc\\&=(a+b)^2+(b+c)^2-ab-bc+ca\end{aligned}

I see nothing helpful. Is the identity I am using is right?

2. ## Re: If a^3+b^3+c^3=3abc, then prove that a=b=c.

I think you need a further condition otherwise

a=-2x

b=x

c=x

satisfies your equation for all x.

If you have the condition that a,b,c > 0 then what you can do is let

$a=x+\delta$

$b=x$

$c=x$

and solve for $\delta$ showing it must be equal to zero.

Then by symmetry of the problem the same result applies to $b$ and $c$.

3. ## Re: If a^3+b^3+c^3=3abc, then prove that a=b=c.

Originally Posted by romsek
I think you need a further condition otherwise

a=-2x

b=x

c=x

satisfies your equation for all x.

If you have the condition that a,b,c > 0 then what you can do is let

$a=x+\delta$

$b=x$

$c=x$

and solve for $\delta$ showing it must be equal to zero.

Then by symmetry of the problem the same result applies to $b$ and $c$.
Yes $a, b, c$ are positive numbers. But how do you say $b=c=x$. Please explain a little bit.

4. ## Re: If a^3+b^3+c^3=3abc, then prove that a=b=c.

Originally Posted by NameIsHidden
Yes $a, b, c$ are positive numbers. But how do you say $b=c=x$. Please explain a little bit.
I assumed it and showed that a must also be equal to x.

I suppose my argument does have a hole in it in that a, b, and c could all be distinct.

5. ## Re: If a^3+b^3+c^3=3abc, then prove that a=b=c.

Originally Posted by romsek
I assumed it and showed that a must also be equal to x.

I suppose my argument does have a hole in it in that a, b, and c could all be distinct.
What to do now? Should I take $a=x + \alpha, b=x+\beta, c=x+\gamma$

6. ## Re: If a^3+b^3+c^3=3abc, then prove that a=b=c.

Originally Posted by NameIsHidden
What to do now? Should I take $a=x + \alpha, b=x+\beta, c=x+\gamma$

7. ## Re: If a^3+b^3+c^3=3abc, then prove that a=b=c.

Originally Posted by romsek
But I just don't understand this

$(a-b)^2+(b-c)^2+(c-a)^2=0\Rightarrow a=b=c$.

How did it imply that one. I understand this logically since a,b,c are positive integers and the three squares are also positive. But can it be proved mathematically?

8. ## Re: If a^3+b^3+c^3=3abc, then prove that a=b=c.

Originally Posted by NameIsHidden
But I just don't understand this

$(a-b)^2+(b-c)^2+(c-a)^2=0\Rightarrow a=b=c$.

How did it imply that one. I understand this logically since a,b,c are positive integers and the three squares are also positive. But can it be proved mathematically?
come on...

in order for the sum to be zero, as all the terms are non-negative, they must all be equal to zero. So

$a-b=0 \Rightarrow a=b$

$b-c=0 \Rightarrow b=c$

$(a=b) \wedge (b=c) \Rightarrow a=c$ so

$a=b=c$

9. ## Re: If a^3+b^3+c^3=3abc, then prove that a=b=c.

Originally Posted by romsek
come on...

in order for the sum to be zero, as all the terms are non-negative, they must all be equal to zero. So

$a-b=0 \Rightarrow a=b$

$b-c=0 \Rightarrow b=c$

$(a=b) \wedge (b=c) \Rightarrow a=c$ so

$a=b=c$
I know that. That's what I said I understand it logically. I thought that there was another way without logic.

10. ## Re: If a^3+b^3+c^3=3abc, then prove that a=b=c.

let us go out in a different way.
Lets see if this criteria is satisfied for 3 cases then it will work.It may not be correct way.

let us consider a=b=c=3

a=b=c=7

a=b=c=5

a=b=c=1

we can see that for 3,5,7 the criteria a3+b3+c3=3abc, will not satisfy for 1 only it will satisfy .
Also for other combinations of a,b,c it will not work.

Originally Posted by NameIsHidden
If $a^3+b^3+c^3=3abc$, then prove that $a=b=c$.

$\displaystyle \text{By the identity, }a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca)\text{ we find either }a+b+c=0\text{ or }a^2+b^2+c^2+ab+bc+ca=0.\text{ the first thing is false since we arrive at a contradiction.}$
\displaystyle \text{ I tried many things even by expressing }\begin{aligned} a^2+b^2+c^2+ab+bc+ca&=a^2+b^2+c^2+2ab+2bc+ca-ab-bc\\&=(a+b)^2+(b+c)^2-ab-bc+ca\end{aligned}

I see nothing helpful. Is the identity I am using is right?

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