In the expansion of ${(3^{-x/4}+3^{5x/4})}^n$  , the sum of the binomial coefficients is 64 and the term with the greatest binomial coefficient exceeds the third by  $(n-1)$ then the value of  $x$ .

Given sum of the binomial coefficients in the expansion of  ${(3^{-x/4}+3^{5x/4})}^n=64$ 
Then putting    $3^{-x/4}=3^{5x/4}=1$
 $\therefore$      ${(1+1)}^n=64$
$\Rightarrow$          $2^n=64$
 $\therefore$             $n=6$
We know that middle term has the greatest binomial coefficient,
Here  $n=6$
$\therefore$  middle term $=(\frac{6}{2}+1)th$ term
 $=4th$    term
     $=T_4$
and given  $T_4=(n-1)+T_3$
$\therefore$     $T_{3+1}=(n-1)+T_{2+1}$
  ${ }^6{C}_3{\left(3^{-x/4}\right)}^3{\left(3^{5x/4}\right)}^3=(6-1){+}^6{C}_2{\left(3^{-x/4}\right)}^4{\left(3^{5x/4}\right)}^2$
   $\Rightarrow \frac{\mathrm{6.5.4}}{\mathrm{1.2.3}}{3}^{-3x/4}{.3}^{15x/4}=5+\frac{6.5}{1.2}{3}^{-x}{.3}^{5x/2}$
   $\Rightarrow {20.3}^{3x}=5+{15.3}^{3x/2}$
Let   $3^{3x/2}=t$      …. (i)   $(t>0)$
$\therefore$       $20t^2-15t-5=0$
   $\Rightarrow 4t^2-3t-1=0$
   $\Rightarrow (4t+1)(t-1)=0$
Binomial Theorem
      $t\ne -\frac{1}{4}$               $(t>0)$
$\therefore$       $t=1$
From (i),  $3^{3x/2}=1=3^0$
   $\therefore 3x/2=0$
   $\therefore x=0$