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(B) The arithmetic mean of $b$ and $c$ is $a$,so $a = \frac{b+c}{2}$,which implies $b+c = 2a$.Given $g_1$ and $g_2$ are two geometric means between $b

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$2$

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(B) The arithmetic mean of $b$ and $c$ is $a$,so $a = \frac{b+c}{2}$,which implies $b+c = 2a$.Given $g_1$ and $g_2$ are two geometric means between $b$ and $c$,the sequence is $b, g_1, g_2, c$ in $G$.$P$.Let the common ratio be $r$. Then $c = b r^3$,so $r^3 = \frac{c}{b}$.$g_1 = br$ and $g_2 = br^2$.Now,$g_1^3 + g_2^3 = (br)^3 + (br^2)^3 = b^3 r^3 + b^3 r^6$.Substituting $r^3 = \frac{c}{b}$:$g_1^3 + g_2^3 = b^3 \left( \frac{c}{b} \right) + b^3 \left( \frac{c}{b} \right)^2 = b^2 c + b^3 \left( \frac{c^2}{b^2} \right) = b^2 c + bc^2 = bc(b+c)$.Since $b+c = 2a$,we have $g_1^3 + g_2^3 = bc(2a) = 2abc$.Comparing this with $kabc$,we get $k = 2$.

The arithmetic mean of two numbers $b$ and $c$ is $a$,and $g_1$ and $g_2$ are two geometric means between them. If $g_1^3 + g_2^3 = kabc$,then $k = \dots$

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