Problem
Solve for a: \[ \sqrt{4+\sqrt{16+16 a}}+\sqrt{1+\sqrt{1+a}}=6 . \] $\left(7 b^{3}\right)^{2} \cdot\left(4 b^{2}\right)^{-3}$, given that bis non-zero.
Solution
Step 1 ：Square both sides of the equation: \(\left(\sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}}\right)^2 = 6^2\)
Step 2 ：Solve for a: \(\boxed{a = 8}\)
