Functions Galore

Similar solution as @Ephram Chun 's

We note that $f(x,y) = (x+y)(x^2 - xy + y^2) = x^3 + y^3$ and that the sum,

$\begin{aligned} S & = f((1,13) + f(6,2) + f(3,10) + f(9,4) + f(8,5) + f(14,7) + f(12,11) \\ & = 1^3 + 2^3 + 3^3 + \cdots + 14^3 \\ & = \sum_{n=1}^{14} n^3 = \left(\sum_{n=1}^{14} n \right)^2 = \left(\frac {14(14+1)}2 \right)^2 = 105^2 = \boxed{11025} \end{aligned}$

$f(x,y)= (x+y)(x^2-xy+y^2)= x^3+y^3$ $f(1,13)+f(6,2)+\cdots+f(12,11)= 1^3+2^3+\cdots+14^3$ $= \left (\dfrac{14.15}{2}\right)^2=105^2=11025$

We can simplify $f(x,y)=(x+y)(x^2-xy+y^2)$ into $f(x,y)=x^3+y^3.$ Let's break down each of the functions now. $f(1,13)=1^3+13^3, f(6,2)=6^3+2^3, f(3,10)=3^3+10^3, f(9,4)=9^3+4^3, f(8,5)=8^3+5^3, f(14,7)=14^3+7^3, f(12,11)=12^3+11^3$ We can see that when we add all of them up we get $1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3+10^3+11^3+12^3+13^3+14^3$ We can use the formula that states $1^3+2^3+3^3+4^3+n^3=(1+2+3+4+...+n)^2$ Therefore the answer is $(1+2+3+4+5+6+7+8+9+10+11+12+13+14)^2=105^2=\boxed{11025}$