What's the best way to compute $\frac{a^4 + b^4 + c^4}{a^2 + b^2 + c^2}$

What you are looking at is $$\frac{1+x^4+(x+1)^4}{1+x^2+(x+1)^2}$$ where $x=2012$. Expanding and simplifying, this is $$\frac{2+4x+6x^2+4x^3+2x^4}{2+2x+2x^2}=\frac{1+2x+3x^2+2x^3+x^4}{1+x+x^2}$$ $$=\frac{(1+x+x^2)^2}{1+x+x^2}=1+x+x^2.$$

Thus the final answer is $$2012^2+2012+1,$$ and from here you can calculate the number by hand.

Hint The given triple $(1, 2012, 2013)$ has the special feature that $1 + 2012 = 2013$. This motivates writing the quotient as $$\frac{a^4 + b^4 + (a + b)^4}{a^2 + b^2 + (a + b)^2}$$ for $a = 1, b = 2012$.

Additional hint Expanding the numerator, we can see that it factors as $2(a^2+ab+b^2)^2$

Hint:

$$\frac{1^4+x^4+(x+1)^4}{1^2+x^2+(x+1)^2}=x^2+x+1$$