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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.

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

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

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