jmc

We have that $$\begin{array}{rl}f(f(f(x)))& =f(f(ax+b))\\ & =f(a(ax+b)+b)=f(a^{2}x+ab+b)\\ & =a(a^{2}x+ab+b)+b\\ & =a^{3}x+a^{2}b+ab+b\\ & =8x+21.\end{array}$$Matching coefficients, we get $a^3=8$ and $a^{2}b+ab+b=21.$ Then $a=2,$ so $4a+2b+b=21,$ or $7b=21,$ so $b=3$.

Therefore, $a+b=\boxed{5}.$