Ta có: \(a_2^2=a_1\cdot a_3\)
=>\(\frac{a_1}{a_2}=\frac{a_2}{a_3}\)
\(a_3^2=a_2\cdot a_4\)
=>\(\frac{a_2}{a_3}=\frac{a_3}{a_4}\)
=>\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}\)
Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=k\)
=>\(\begin{cases}a_3=a_4\cdot k\\ a_2=a_3\cdot k=a_4\cdot k^2\\ a_1=a_2\cdot k=a_4\cdot k^3\end{cases}\)
\(\frac{a_1^3+a_2^3+a_3^3}{a_2^3+a_3^3+a_4^3}=\frac{\left(a_4\cdot k^3\right)^3+\left(a_4\cdot k^2\right)^3+\left(a_4\cdot k\right)^3}{\left(a_4\cdot k^2\right)^3+\left(a_4\cdot k\right)^3+a_4^3}\)
\(=\frac{a_4^3\cdot k^3\left(k^6+k^3+1\right)}{a_4^3\left(k^6+k^3+1\right)}=k^3\)
\(\frac{a_1}{a_4}=\frac{a_4\cdot k^3}{a_4}=k^3\)
Do đó: \(\frac{a_1^3+a_2^3+a_3^3}{a_2^3+a_3^3+a_4^3}=\frac{a_1}{a_4}\)
