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As stated in the title, I'm supposed to show that $(a+b+c)^3 = a^3 + b^3 + c^3 + (a+b+c)(ab+ac+bc)$.

My reasoning: $$(a + b + c)^3 = [(a + b) + c]^3 = (a + b)^3 + 3(a + b)^2c + 3(a + b)c^2 + c^3$$

$$(a + b + c)^3 = (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + b^2)c + 3(a + b)c^2+ c^3$$

$$(a + b + c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$$

$$(a + b + c)^3 = (a^3 + b^3 + c^3) + (3a^2b + 3a^2c + 3abc) + (3ab^2 + 3b^2c + 3abc) + (3ac^2 + 3bc^2 + 3abc) - 3abc$$

$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3a(ab + ac + bc) + 3b(ab + bc + ac) + 3c(ac + bc + ab) - 3abc$$

$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3(a + b + c)(ab + ac + bc) - 3abc$$

$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3[(a + b + c)(ab + ac + bc) - abc]$$ It doesn't look like I made careless mistakes, so I'm wondering if the statement asked is correct at all.

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17  
The statement is obviously wrong: look at it when $a=b=c=1$. – KCd Jan 28 '13 at 14:43
    
You've probably not mentioned some condition? – hjpotter92 Jan 28 '13 at 14:45
    
I transcribed the exercise. Found a typo or another at other parts of the book, so that's just another one of them. – Sawyier Jan 28 '13 at 14:48
    
If it helps the final statement in your derivation looks correct. – kaine Jan 28 '13 at 15:12
1  
Incorrect questions teach you to figure out if a statement is true before wasting time trying to prove it. – DanielV Jul 20 '14 at 2:58
up vote 7 down vote accepted

In general, $$a^n+b^n+c^n = \sum_{i+2j+3k=n} \frac{n}{i+j+k}\binom {i+j+k}{i,j,k} s_1^i(-s_2)^js_3^k$$

where $s_1=a+b+c$, $s_2=ab+ac+bc$ and $s_3=abc$ are the elementary symmetric polynomials.

In the case that $n=3$, the triples possible are $(i,j,k)=(3,0,0),(1,1,0),$ and $(0,0,1)$ yielding the formula:

$$a^3+b^3+c^2 = s_1^3 - 3s_2s_1 + 3s_3$$

which is the result you got.

In general, any symmetric homogeneous polynomial $p(a,b,c)$ of degree $n$ can be written in the form:

$$p(a,b,c)=\sum_{i+2j+3k=n} a_{i,j,k} s_1^i s_2^j s_3^k$$

for some constants $a_{i,j,k}$.

I've often thought Fermat's Last Theorem was most interesting when stated as a question about these polynomials. One statement of Fermat can be written as:

If $p$ is an odd prime, then $a^n+b^n+c^n=0$ if and only if $a+b+c=0$ and $abc=0$.

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Regarding the last statement, shouldn't the three $n$s in the exponents be $p$s? – wythagoras Jan 30 at 18:58

$(a+b+c)^3 + 3abc = a^3 + b^3 + c^3 + (a+b+c)(ab+ac+bc)$, that's the right factorization I guess

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You have a missing factor of $3$ on the right. – Start wearing purple Sep 9 '13 at 23:11

$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$ with $3(a+b)(b+c)(c+a) = 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$

I guess the right factorization

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protected by John Ma Oct 17 '15 at 4:20

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