(a) a, b, c are distinct real numbers such that $$a^3 = 3(b^2+c^2)-25$$, $$b^3=3(c^2+a^2)-25$$, $$c^3=3(a^2+b^2)-25$$. Find the numerical value of $$abc$$.
(b) $$a = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{2015^2}$$,
find $$[a]$$, where $$[a]$$ denotes the integer part of a.

For part (a), subtracting pairs of the given equations and factoring gives $$a+b+c=-3$$; substituting this into the sum of all three original equations and using the identity for $$a^3+b^3+c^3$$ eventually isolates $$abc = 2-7(ab+bc+ca)$$, and a second combination of the equations gives $$ab+bc+ca=0$$, so $$abc=2$$. For part (b), bounding each term $$\frac{1}{n^2}$$ between $$\frac{1}{n-1}-\frac{1}{n}$$ and telescoping shows the sum $$\frac{1}{2^2}+\cdots+\frac{1}{2015^2}$$ is strictly less than 1 (specifically close to but under 1), so $$a = 1 + (\text{something less than }1)$$, giving $$[a]=1$$.