Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
11  
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
    
Incorrect questions teach you to figure out if a statement is true before wasting time trying to prove it. –  DanielV Jul 20 at 2:58

4 Answers 4

up vote 4 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 $abc=0$.

share|improve this answer

Take $a=b=c=1$, you will find that that the expression is not correct.

share|improve this answer

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

share|improve this answer
    
You have a missing factor of $3$ on the right. –  O.L. Sep 9 '13 at 23:11

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

I guess the right factorization

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.