Two infinite radicals question Hello I have stucked with theese two questions:
$\sqrt{a:\sqrt{a:\sqrt{a: \cdots}}} + \sqrt[3]{a\cdot\sqrt[3]{a\cdot\sqrt[3]{a\cdots}}} = 12$
$a=\text{ ?}$

$\sqrt{6+\sqrt{7+\sqrt{6-\sqrt{6-\sqrt{6- \cdots}}}}}=?$

$\sqrt{a/\sqrt{a/\sqrt{a/ ...}}}=a^\frac{1}{2}/(a^{\frac{1}{4}}/(a^{\frac{1}{8}}...))=a^{\sum_{i=1}^{\inf}\frac{1}{2*(-2)^{i-1}}}=a^{\frac{1}{3}} $
$\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a ...}}}=a^{\sum_{i=1}^{inf}\frac{1}{3i}}=a^{\frac{1}{2}}
$
Then we have $a^{\frac{1}{3}}+a^{\frac{1}{2}}=12$
Function $f(a)=a^{\frac{1}{3}}+a^{\frac{1}{2}}$ is strictly increasing so equation can't have more than 1 solution. It's easy to see that solution is 64.
For second part let $t=\sqrt{6-\sqrt{6-..}}$
But then $ t^2=6-\sqrt{6-\sqrt{6-..}}=6-t  $
$t^2+t-6=0$
Using t>0 we have t=2
then we get $\sqrt{6+\sqrt{7+t}}=\sqrt{6+\sqrt{9}}=\sqrt{6+3}=3 $
