An algebra problem by A Former Brilliant Member

$(i): a = \frac{b+c}{2}$

$(ii): \frac{c}{g_2} = \frac{g_2}{g_1} = \frac{g_1}{b}$

$(iia): g_2^2 = c\cdot g_1$

$(iib): g_1^2 = b\cdot g_2$

Plugging $(iia)$ in $(iib)$ :

$g_2^4 = b\cdot c^2\cdot g_2$

$(iii): g_2^3 = b\cdot c^2$

Now, plugging $(iii)$ in $(iia)$ :

$g_1 = \frac{g_2^2}{c}$

$g_1^3 = \frac{g_2^6}{c^3}$

$g_1^3 = \frac{b^2\cdot c^4}{c^3}$

$(iv): g_1^3 = b^2\cdot c$

Adding $(iii)$ and $(iv)$ :

$g_1^3 + g_2^3 = b^2\cdot c + b\cdot c^2 = bc(b+c) = bc\cdot 2a = \fbox{2abc}$

$b, g_1 , g_2 , c$ are in GP, let $r$ be their common ratio

$r = \begin{pmatrix} \dfrac cb \end{pmatrix} ^ {\frac 13}$

Also, $a = \dfrac 12 (b+c)$

Now, $g_1^3 + g_2^3 = b^3( r^3 + r^6) = b^3\begin{pmatrix} \dfrac cb + \begin{pmatrix} \dfrac cb \end{pmatrix} ^2 \end{pmatrix} = b^3 \dfrac cb \begin{pmatrix} \dfrac {b+c}{b} \end{pmatrix} = bc \times 2 \times \dfrac {(b+ c)}{2} = \boxed {2abc}$