Two infinite radicals question

Question

Hello I have stucked with theese two questions:

$\sqrt{a:\sqrt{a:\sqrt{a: \cdots}}} + \sqrt[3]{a\cdot\sqrt[3]{a\cdot\sqrt[3]{a\cdots}}} = 12$

$a=\text{ ?}$

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

Answer 1

$\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 $