# Intuition behind problem in (classical) algebra

To give some background, the question is to show that if $a=b+c$ then $$a^4+b^4+c^4 = 2a^2b^2+2b^2c^2+2c^2a^2$$

Which, for completeness, I was able to do by squaring twice $$(a-b-c)^2=0$$ gives $$a^2+b^2+c^2= 2(bc-ac-ab)$$ which on squaring gives the required identity. $$a^4+b^4+c^4 +2a^2b^2+2b^2c^2+2c^2a^2 = 4a^2b^2+4b^2c^2+4c^2a^2 + 8(a^2bc - ab^2c-abc^2)$$ $$a^4+b^4+c^4 =2a^2b^2+2b^2c^2+2c^2a^2 +8abc(a-b-c)$$

$$a^4+b^4+c^4 = 2a^2b^2+2b^2c^2+2c^2a^2$$

For the second part the statement is: "explain why an unsymmetric equation gives such a symmetric expression." I ignored this part while solving it as I did not have a very good impression of the author of the problem set from the previous problems and as no rigorous definition of symmetry was given the author was presumably requiring some kind of aesthetic appreciation.

The reason as it turns out in the hint is that "on squaring twice effect of minus signs gets cancelled." If this statement is taken for granted, from this logic all the ring permutations of ($\pm a,\pm b,\pm c$),i.e 4 yield the same equation. It is indeed so as the factors turn out to be $$(a+b+c)(a-b-c)(a+b-c)(a-b+c)$$

However, by this same logic $(a+b+c)^4=(a-b+c)^4$ But it is not. The reason I am asking here is that I can verify the above algebraically and see why everything is ok, but cannot explain it as the way algebra works for me is that I work it out all keeping care for all the minus signs then magically half the terms seem to cancel out and I am left with a single expression out of the blue.

Sorry for the wall of text.

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I'm sorry, what is the question? – Gerry Myerson Jul 7 '11 at 5:58
@Gerry How does "squaring twice cancel the effect of the minus sign", yielding a symmetric equation and does not in the other cases. – kuch nahi Jul 7 '11 at 6:00

To bring out the symmetry, let $x=a$, $y=-b$, $z=-c$. You are then looking at the equation $x+y+z=0$.

Added: This is a small sample from a rich subject that, apart from simple cases, began with Newton. The issue is the relationship between the elementary symmetric functions in $n$ variables and sums of powers. Look for material under Newton Identities.

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Expanding the area of the triangle $ABC$ of sides $a,b,c$ by Heron's formula, you'll get

$$Area(ABC)=\frac{1}{16} (2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4)$$

If $a=b+c$ then the triangle is degenerate and has area zero. Therefore

$$2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4=0$$

holds.

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You must have been involved in contest training. – André Nicolas Jul 7 '11 at 7:02
Neat, but I suppose OP had discovered this himself when he factored the expression. But +1 for pointing this out :-) – Aryabhata Jul 7 '11 at 7:02
If the OP had discovered this, then the intuition would be very clear. Anyway, this is for geometric intuition, not algebraic. :) – Beni Bogosel Jul 7 '11 at 7:05

You can write the fourth power as $$(\epsilon_x x + \epsilon_y y + \epsilon_z z)^4,$$ where in your case $\epsilon_x = 1$ and $\epsilon_y = \epsilon_z = -1$. You then get some expression involving $(\epsilon_x x)^2,(\epsilon_y y)^2,(\epsilon_z z)^2$. Since $(\pm 1)^2 = 1$, it should make any difference whether you choose a plus sign or minus sign.

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Thank you. This reinforces the above answer in more detail. I was looking at the problem in reverse. – kuch nahi Jul 7 '11 at 6:09
To add something though, it is that we get an expression in which the lowest power is $2$ that is why it is valid, otherwise, $(x+y+z)^4\neq (x-y+z)^4$ – kuch nahi Jul 7 '11 at 6:12