# Larson 4.1.10

1. Jan 7, 2008

### ehrenfest

[SOLVED] Larson 4.1.10

1. The problem statement, all variables and given/known data
Determine all triplets of integers (x,y,x) satisfying the equation

$$x^3+y^3+z^3=(x+y+z)^3$$

2. Relevant equations

3. The attempt at a solution

I think the only solutions are x=y=0, x=z=0, z=y=0, z=-y and x=0, x=-z and y=0, x=-y and z=0. If I rewrite the equation as (x+y)(x+z)(y+z)=-xyz, several things are obvious:
1)x,y,z are not all odd
2)x,y,z are not all positive and not all negative
3)if for example, z is negative and x,y are positive, then abs(z)< x,y or abs(z)>x,y

But how can I show there are no other solutions...

2. Jan 7, 2008

certainly either all are 0 or none
by fermat last theorm(diff. from earliar one)

i will work it out some other time

3. Jan 7, 2008

### ehrenfest

I am sure there is a way to do this without using Fermat's Last Theorem. None of Larson's problems require something that advanced.

4. Jan 7, 2008

### morphism

I don't even see how Fermat is relevant here.

I have one comment on your attempted solution:
That doesn't look correct to me. Shouldn't the right side be simply 0?

5. Jan 7, 2008

### ehrenfest

No. Multiply it out. The LHS only gives 2xyz.

6. Jan 7, 2008

### morphism

Maybe it's the sleepiness, but it seems to me that (x+y+z)^3 = x^3 + y^3 + z^3 + 3(x+y)(x+z)(y+z).

7. Jan 7, 2008

### ehrenfest

You're right. It does. The solutions to (x+y)(x+z)(y+z)=0 (and to the the original equation) are x=-y,y=-z, or z=-x.

Last edited: Jan 7, 2008