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(B) Given that $a$ is the arithmetic mean of $b$ and $c$,we have $a = \frac{b+c}{2}$,which implies $b+c = 2a$.Since $G_1$ and $G_2$ are two geometric

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$2 G_1 G_2 a$

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(B) Given that $a$ is the arithmetic mean of $b$ and $c$,we have $a = \frac{b+c}{2}$,which implies $b+c = 2a$.Since $G_1$ and $G_2$ are two geometric means between $b$ and $c$,the sequence $b, G_1, G_2, c$ is in $G$.$P$.Let the common ratio be $r$. Then $G_1 = br$,$G_2 = br^2$,and $c = br^3$.Thus,$r = (c/b)^{1/3}$.We know $G_1 G_2 = (br)(br^2) = b^2 r^3 = b^2 (c/b) = bc$.Now,$G_1^3 + G_2^3 = (br)^3 + (br^2)^3 = b^3 r^3 + b^3 r^6 = b^3 (c/b) + b^3 (c/b)^2 = b^2 c + b c^2 = bc(b+c)$.Substituting $bc = G_1 G_2$ and $b+c = 2a$,we get:$G_1^3 + G_2^3 = (G_1 G_2)(2a) = 2 G_1 G_2 a$.

If $a$ is the arithmetic mean of $b$ and $c$ and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $

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