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

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 $a_{i,j,k}$.

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 $abc=0$.

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