Đáp án:
$\begin{array}{l}
a_2^2 = {a_1}.{a_3} \Rightarrow \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}\\
a_3^2 = {a_2}.{a_4} \Rightarrow \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{{a_4}}}{{{a_3}}}\\
\Rightarrow \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{{a_4}}}{{{a_3}}}\\
\text{Đặt}:\dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{{a_4}}}{{{a_3}}} = k\\
\Rightarrow \left\{ \begin{array}{l}
{a_2} = {a_1}.k\\
{a_3} = {a_2}.k\\
{a_4} = {a_3}.k
\end{array} \right.\\
\Rightarrow {a_4} = k.\left( {{a_2}.k} \right) = {k^2}.{a_2} = {k^2}.{a_1}.k = {a_1}.{k^3}\\
\Rightarrow \dfrac{{{a_1}}}{{{a_4}}} = \dfrac{{{a_1}}}{{{a_1}.{k^3}}} = \dfrac{1}{{{k^3}}}\\
\dfrac{{a_1^3 + a_2^3 + a_3^3}}{{a_2^3 + a_3^3 + a_4^3}} = \dfrac{{a_1^3 + a_2^3 + a_3^3}}{{{{\left( {{a_1}.k} \right)}^3} + {{\left( {{a_2}.k} \right)}^3} + {{\left( {{a_3}.k} \right)}^3}}}\\
= \dfrac{{a_1^3 + a_2^3 + a_3^3}}{{\left( {a_1^3 + a_2^3 + a_3^3} \right).{k^3}}} = \dfrac{1}{{{k^3}}}\\
\text{Vậy}\,\dfrac{{a_1^3 + a_2^3 + a_3^3}}{{a_2^3 + a_3^3 + a_4^3}} = \dfrac{{{a_1}}}{{{a_4}}}\left( { = \dfrac{1}{{{k^3}}}} \right)
\end{array}$
