Function Machine

First, find what $ax+b$ is when $f\left( x \right)$ becomes $f\left( f\left( f\left( x \right) \right) \right)$ . $f\left( f\left( f\left( x \right) \right) \right) =a\left( a\left( ax+b \right) +b \right) +b\\ f\left( f\left( f\left( x \right) \right) \right) ={ a }^{ 3 }x+{ a }^{ 2 }b+ab+b$ Since $x$ only appears once in the equation, we can determine from the given function, $f\left( f\left( f\left( x \right) \right) \right) =8x+21$ , that ${ a }^{ 3 }=8$ . Hence, $a=2$ . If you plug $a$ back into the most recent function, you get $f\left( f\left( f\left( x \right) \right) \right)= 8x+4b+2b+b =8x+7b$ Looking back at the given function, we can determine that $7b=21\Longrightarrow b=3$ Therefore, $a+b=\boxed{5}$ .

First we expand the function $f(f(f(x)))$

We know $f(x) = ax + b$ so we can substitute that value in

$f(f(f(x))) = f(f(ax+b))$

From here we expand and simplify

$f(f(ax + b)) = f(a(ax + b) + b)$

$f(a(ax + b) + b) = a(a(ax + b) + b) + b$

$a(a(ax + b) + b) + b = a^{2}(ax + b) + ba + b$

$a^{2}(ax + b) + ba + b = a^{3}x + ba^{2} + ba + b$

$a^{3}x + ba^{2} + ba + b = a^{3}x + b(a^{2} + a + 1)$

$a^{3}x + b(a^{2} + a + 1) = 8x + 21$

Next we separate the $x$ from both equations and put it in one of it's own.

$a^{3}x = 8x$

$b(a^{2} + a + 1) = 21$

Next we simplify the first equation

$a^{3} = 8$

$a = \sqrt[3]{8}$

$a = 2$

We then substitute that into the other equation

$b((2)^{2} + (2) + 1) = 21$

$7b = 21$

$b = \frac{21}{7}$

$b = 3$

Now that we know $a$ and $b$ we can calculate $a + b$

$a + b = (2) + (3) = 5$

$\boxed{a + b = 5}$