Number theory using algebraic identities

$\begin{aligned} Q & = \frac {2001^4+2000^4+1^4}{2001^2+2000^2+1^2} & \small \blue{\text{Let }a = 2000} \\ & = \frac {(a+1)^4+a^4+1}{(a+1)^2+a^2+1} \\ & = \frac {a^4+4a^3+6a^2+4a+1+a^4+1}{a^2+2a+1+a^2+1} \\ & = \frac {2a^4+4a^3+6a^2+4a+2}{2a^2+2a+2} \\ & = \frac {a^4+2a^3+3a^2+2a+1}{a^2+a+1} \\ & = a^2+a+1 & \small \blue{\text{Putting back }a = 2000} \\ & = 2000^2 + 2000+1 \\ & = 4002001 \end{aligned}$

Therefore the sum of digits of $Q$ is $4+2+1 = \boxed 7$ .