# Homework Help: 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)

21. Jan 6, 2013

### MrWarlock616

I think what joffan said would be easiest! :)

22. Jan 6, 2013

### Curious3141

OK, I'll post this hint. This method is essentially identical to what haruspex posted in post #8, but more explicit. Micromass reiterated the same solution in post #16, but it's easier to "grasp" with the omega symbol than with the exponential notation.

There are three cube roots of unity, $1, \omega, \omega^2$.

The original system can be reduced to:

$a + b = \omega(b + c)$

$b + c = \omega(c + a)$, which is equivalent to $a + b = \omega^2 (c + a)$

and these are solved simultaneously.

The system of equations has to be just so, because:

$a + b = (1)(b + c)$ would imply that a = c, violating uniqueness of a, b and c.

and

$a + b = \omega(c + a)$ would imply that $\omega (b + c) = \omega (c + a)$, giving a = b, again violating uniqueness.

So that's the only admissible system of equations, and they can easily be solved by substitution. Then using the well-worn algebraic properties of $\omega$ and $\omega^2$, e.g. $\omega^2 = -1 - \omega$ and $(1 + \omega)^3 = -1$, etc., the solutions can easily be manipulated to prove $a^3 = b^3 = c^3$.

Do you understand the outline of this solution? It's a lot less tedious than going the direct algebraic route.

Last edited: Jan 6, 2013
23. Jan 6, 2013

### HallsofIvy

The point is to do something yourself! If you do not know how to solve a problem try many different ways to see if one works or if the result of trying suggest something else. If you are given different ways to solve a problem and don't know which is "best", try one and see if it works. Even if it is not the best way, you will have solved the problem which is the entire point!

24. Jan 6, 2013

### Hurkyl

Staff Emeritus
The homogeneous equation (a+b)^3 = (a+c)^3 defines an algebraic curve of degree 3 in the projective plane (with indeterminates a, b, c).

So does the homogeneous equation (a+b)^3 = (b+c)^3.

Therefore, exactly one of the following two statements is true:

• The two curves have a common component, and thus there are infinitely many solutions
• The two curves intersect in 3*3 = 9 points in the complex projective plane, counting multiplicity.

We can check that the following gives 9 distinct projective solutions:
$$a = R \omega^m \qquad b = R \omega^n \qquad c = R$$
where $\omega$ is a cube root of unity, R is a nonzero complex number, and m,n are integers. (projective solutions ignore scaling, so R=1 and R=2 would both give the same projective solution if m,n are kept the same)

If we can show that there aren't infinitely many solutions, then this, along with a=b=c=0, must be the complete solution set to the original equation.

25. Jan 6, 2013

### VantagePoint72

Hurkyl, given the level at which the question is posed, do you really think even your first sentence is going to be intelligible to him or her? I don't see what the point is in providing a solution—even if it's correct—that is obviously enormously beyond the asker's academic level.