Lời giải:Ta có \({{a}^{{{\log }_{b}}a}}+{{16}^{{{\log }_{a}}\left( \frac{{{b}^{8}}}{{{a}^{3}}} \right)}}=12{{b}^{2}}\Leftrightarrow {{a}^{{{\log }_{b}}a}}+16{{b}^{8{{\log }_{a}}b-3}}=12{{b}^{2}}.\)
Đặt \(t={{\log }_{b}}a\Leftrightarrow a={{b}^{t}}\) và \({{\log }_{a}}b=\frac{1}{t}.\) Do đó \(\left( * \right)\Leftrightarrow {{a}^{t}}+16{{b}^{\frac{8}{t}-3}}=12{{b}^{2}}\)
\(\Leftrightarrow 12{{b}^{2}}={{b}^{{{t}^{2}}}}+8{{b}^{\frac{8}{t}-3}}+8{{b}^{\frac{8}{t}-3}}\ge 3\sqrt[3]{{{b}^{{{t}^{2}}}}.8{{b}^{\frac{8}{t}-3}}.8{{b}^{\frac{8}{t}-3}}}=12\sqrt[3]{{{b}^{{{t}^{2}}+\frac{8}{t}+\frac{8}{t}-6}}}\)
Suy ra \({{a}^{t}}+16{{b}^{\frac{8}{t}-3}}\ge 12{{b}^{2}}\). Dấu bằng xảy ra khi
\(\left\{ {\begin{array}{*{20}{l}}
{{t^2} = \frac{8}{t}}\\
{{b^{{t^2}}} = 8{b^{\frac{8}{t} - 3}}}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{t = 2}\\
{{b^4} = 8b}
\end{array}} \right. \Leftrightarrow b = 2\)
Mà \(a={{b}^{t}}={{2}^{2}}=4\to \to {{a}^{3}}+{{b}^{3}}={{2}^{3}}+{{4}^{3}}=72.\)
