Let a and b are two positive real numbers such that $a^2+b=2$ then the maximum value of term independent of x in the expansion ${(a{x}^{1/6}+b{x}^{-1/3})}^9$ is 

 $Tr+1={}^9{c}_r{a}^{9-r}.b^r.x^{\frac{3-r}{2}}$
For independent term r = 3
Independent term =  ${}^9{c}_3{a}^6{b}^3{=}^9{c}_3$
For max value of  $a^6{b}^3=A.M\ge G.M$
$\Rightarrow \frac{3(a^2/3)+3(b/3)}{6}\ge {\left(\frac{a^6{b}^3}{36}\right)}^6$
$\Rightarrow$ max of  $a^6{b}^3=1$