jmc

Perhaps the fastest way is to use the difference of squares factorization: $$\begin{array}{rl}(a^2+b{)}^2-(a^2-b{)}^2& =[(a^2+b)+(a^2-b)]\cdot [(a^2+b)-(a^2-b)]\\ & =(a^2+b+a^2-b)\cdot (a^2+b-a^2+b)\\ & =(2a^2)\cdot (2b)\\ & =4a^{2}b.\end{array}$$Since $a=4$ and $b=1$, this last expression is equal to $$4\cdot 4^2\cdot 1=4\cdot 16=\boxed{64},$$so that is our answer.

We could also plug in the values of $a$ and $b$ right away and then expand. We then get $$\begin{array}{rl}(a^2+b{)}^2-(a^2-b{)}^2& =(4^2+1{)}^2-(4^2-1{)}^2\\ & =(16+1{)}^2-(16-1{)}^2\\ & =17^2-15^2.\end{array}$$Now, $17^2=289$, and $15^2=225$, so our answer is then $$289-225=89-25=64,$$as before.