Show that $(a+b+c)^3 = a^3 + b^3 + c^3+ (a+b+c)(ab+ac+bc)$

Question

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.

The statement is obviously wrong: look at it when $a=b=c=1$. — KCd, Jan 28 '13 at 14:43
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 '14 at 02:58
Relevant: https://en.wikipedia.org/wiki/Newton%27s_identities — DVD, Nov 10 '18 at 05:06
How does OP have 5 questions and 5 Famous Question badges? Is this a joke? — IraeVid, Jun 22 '23 at 15:39

Thomas Andrews · Accepted Answer · 2021-08-18T18:29:14.843

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:

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

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 $a_{i,j,k}$.

If you don’t know the first statement, you can deduce the values $a_{i,j,k}$ by solving linear equations.

This is because the triples $(i,j,k)$ are limited to $(3,0,0), (1,1,0),(0,0,1).$. So if:

$$a^3+b^3+c^3=a_{3,0,0}(a+b+c)^3+a_{1,1,0}(ab+ac+bc)(a+b+c)+a_{0,0,1}abc$$

Then try it for specific values of $(a,b,c).$ For example, when $(a,b,c)=(1,0,0),$ you get:

$$1=a_{3,0,0}\cdot 1+a_{1,1,0}\cdot 0+a_{0,0,1}\cdot 0.$$

Try $(a,b,c)=(1,1,0)$ and $(1,1,1).$

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 $a^p+b^p+c^p=0$ if and only if $a+b+c=0$ and $abc=0$.

Regarding the last statement, shouldn't the three $n$s in the exponents be $p$s? — wythagoras, Jan 30 '16 at 18:58

score 5 · Answer 2 · edited Jan 05 '19 at 19:17

5

$(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b+c)(ab+ac+bc) - 3abc$ is the right factorisation

edited Jan 05 '19 at 19:17

Confuse

1,100

answered Sep 09 '13 at 22:44

Carlos Peter

51
1
1

You have a missing factor of $3$ on the right. – Start wearing purple Sep 09 '13 at 23:11

score 1 · Answer 3 · edited Jan 26 '15 at 17:36

1

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

edited Jan 26 '15 at 17:36

Debanjan Basu

253

answered Jul 20 '14 at 01:51

Quang Vũ

11

Show that $(a+b+c)^3 = a^3 + b^3 + c^3+ (a+b+c)(ab+ac+bc)$

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