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

The correct option is B.$2abc$

Since, $a$ is the arithmetic mean of $b$ and $c$, therefore,

$\Rightarrow 2a=b+c$...(i)

Now, According to the question, the sequence so formed is,

$\Rightarrow b,G_1,G_2,c$

$=b,br,b{r}^2,b{r}^3$ where $r$ is the common ratio

Thus, $G_1=br,G_2=b{r}^2$ and $c=b{r}^3$

Consider $c=b{r}^3$

$\Rightarrow r^3=\frac{c}{b}$...(ii)

Now, consider, $G_1^3+G_2^3$,

$=(br{)}^3+(b{r}^2{)}^3$

$=b^3{r}^3+b^3{r}^6$

$=b^3{r}^3(1+r^3)$

$=b^3\left(\frac{c}{b}\right)(1+\frac{c}{b})$ [From (ii)]

$=b^{2}c\left(\frac{b+c}{b}\right)$

$=bc\left(2a\right)$ [From (i)]

$=2abc$