If f(x)=ax+b and f(f(f(x)))=8x+21 and if a,b are real numbers then a+…

\(f(x)=ax+b\) so $f(f(x))=(a(ax+b))+b$

$=a^2x+ab+b$

$f(f(f(x)))=a^2(ax+b)+ab+b$

$=a^3x+a^2b+ab+b=8x+21$

so $a^3=8⇒a=2$

so $a^2b+ab+b=21$

$⇒4b+2b+b=21⇒b=3$

$a+b=5$