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Solution

Given $a = \dfrac{b + c}{2}$ and $G_1, G_2$ are geometric means between $b$ and $c$. So $b, G_1, G_2, c$ are in G.P. Let common ratio = $r$. Then $G_1 = br$ and $G_2 = br^2$, $c = br^3$. Hence $a = \dfrac{b + c}{2} = \dfrac{b + br^3}{2} = \dfrac{b(1 + r^3)}{2}.$ Now, $G_1^3 + G_2^3 = b^3(r^3 + r^6) = b^3r^3(1 + r^3).$ But $c = br^3$ and $(1 + r^3) = \dfrac{2a}{b}$. So, $G_1^3 + G_2^3 = b^3r^3 \times \dfrac{2a}{b} = 2ab^2r^3 = 2abc.$