An Intriguing Sum

Given that $a=999$ , $b=333$ , and $c=111$ , implies that $a=3b=9c$ .

$\begin{aligned} S & = 1 + a + 3b + 9c + 3ab + 9ac + 27bc+27abc \\ & = 1 + a + a + a + a^2 + a^2 + a^2 + a^3 \\ & = 1 + 3a + 3a^2 + a^3 \\ & - (1+a)^3 = (1+999)^3 = 1000^3 \\ & = \boxed{1000 000 000} \end{aligned}$

Notice that $b=\frac{a}{3}$ and $c=\frac{a}{9}$ . Substituting in to the formula yields,

$1+a+3\frac{a}{3}+9\frac{a}{9}+3a\frac{a}{3}+9a\frac{a}{9}+27\frac{a}{3}\frac{a}{9}+27a\frac{a}{3}\frac{a}{9}$

$=1+a+a+a+a^{2}+a^{2}+a^{2}+a^{3}$

$=1+3a+3a^{2}+a^{3}$

Notice that coefficients make up the 4th row of Pascal's Triangle, meaning this expression factors into the cube of a binomial. Thus, we can write,

$1+3a+3a^{2}+a^{3}=(1+a)^{3}$

We can now substitute in the given value of $a=999$ to find $(1+999)^{3}=(1000)^{3}=1,000,000,000$