# Different complex numbers

1. Jan 4, 2013

### redount2k9

We have a,b,c different complex numbers so

(a+b)^3 = (b+c)^3 = (c+a)^3

Show that a^3 = b^3 = c^3

From the first equality I reached a^3 - c^3 + 3b(a-c)(a+b+c) = 0 How a is different from c => a-c is different from 0
How do I show that a^3 - c^3 = 0?

2. Jan 4, 2013

### MrWarlock616

If you can prove that 3b(a-c)(a+b+c)=0, you're done.
a, b, c are complex numbers.

3. Jan 4, 2013

### redount2k9

But how to prove that?

4. Jan 4, 2013

### MrWarlock616

Well, you can try assigning a value to the complex numbers and put them in the expression.

5. Jan 4, 2013

### micromass

Huh? How can that ever be a good proof??

6. Jan 4, 2013

### MrWarlock616

lol i think it can't be..but you can do it if you take the time..
edit: expand the expression first.

Last edited: Jan 4, 2013
7. Jan 4, 2013

### haruspex

That's a good start. (It will turn out that a+b+c=0.) If you take out the factor a-c, and also write down a corresponding equation with a/b/c rotated around, then add the two, what do you get?

8. Jan 4, 2013

### haruspex

Doh! There's a much easier way.
We know (a+b) = (b+c)ω, where ω3=1. If a≠c then ω≠1. Writing the corresponding eqn for c+a versus b+c, and assuming b≠a and b≠c, we have (b+c) = (c+a)ω. (If (b+c) = (c+a)ω2 then b=c.) It's not hard from there.

9. Jan 5, 2013

### MrWarlock616

$x^3-y^3=(x-y)(x^2+xy+y^2)$

You will need to prove 3b(a-c)(a+b+c)=0 using this forumla, as I said earlier. Or you can put $a=x_1+iy_1, b=x_2+iy_2, c=x_3+iy_3$, but this will become lengthy.

10. Jan 5, 2013

### Curious3141

11. Jan 5, 2013

### MrWarlock616

12. Jan 5, 2013

### Joffan

I don't know if the original poster is still with us...

We know a, b, and c are distinct.

Therefore (a+b), (b+c) and (c+a) are also distinct

Therefore , as per the question, they are the three distinct cube roots of some number z.

And therefore what is the sum ((a+b) + (b+c) + (c+a))?

13. Jan 5, 2013

### MrWarlock616

2(a+b+c)

14. Jan 5, 2013

### Joffan

More specifically, what is the sum of the three distinct cube roots of some complex number z?

15. Jan 5, 2013

### MrWarlock616

That's 0 ..there you go. This can be locked. :P

16. Jan 5, 2013

### micromass

Anyway. Use that if

$$(a+b)^3=(b+c)^3$$

then

$$a+b=e^{2ik\pi/3}(b+c)$$

for $k\in \mathbb{Z}$. And the same thing for the equality $(a+b)^3=(a+c)^3$.

Now solve the equations

$$\left\{\begin{array}{l} a+b=e^{2ik\pi/3}(b+c)\\ a+b=e^{2ik^\prime\pi/3}(a+c) \end{array}\right.$$

17. Jan 5, 2013

### micromass

Thread is open again. Let's actually wait for the OP to reply before helping any further :tongue2:

18. Jan 6, 2013

### redount2k9

What should I reply? All of you have different opinions... it's hard for me to understand something.

19. Jan 6, 2013

### MrWarlock616

hahahaha use any of the methods..

20. Jan 6, 2013

### redount2k9

Hahahaha why not to use the best method? (if I would know which it is)