Let a,b,c be complex numbers such that:

$\displaystyle \displaystyle{(a+b)(a+c)=b,}$

$\displaystyle \displaystyle{(b+c)(b+a)=c,}$

$\displaystyle \displaystyle{(c+a)(c+b)=a.}$

Prove that $\displaystyle $$a,b,c \in {\Cal R}$$$

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- May 24th 2011, 09:36 PMmathfunComplex Numbers
Let a,b,c be complex numbers such that:

$\displaystyle \displaystyle{(a+b)(a+c)=b,}$

$\displaystyle \displaystyle{(b+c)(b+a)=c,}$

$\displaystyle \displaystyle{(c+a)(c+b)=a.}$

Prove that $\displaystyle $$a,b,c \in {\Cal R}$$$ - May 24th 2011, 10:42 PMProve It
I would start by expanding all three sets of brackets. See what you can do from there...

- May 29th 2011, 12:33 PMrtplol
Let's dissect the question a little bit.

If we expand those brackets, we find:

a^2 + ac + ab + bc = b

b^2 + ab +bc +ac = c

c^2 + bc + ac + ab = a

Note that all 3 equations have the same 3 terms : ab, ac and bc. So let's say "Let x = ab + ac + bc". Then our equations look more like:

a^2 + x = b

b^2 + x = c

c^2 + x = a

Does it become more clear now how to show this?