Find the 9th term of the geometric sequence

$\dfrac{a_{6}}{a_{3}} = \dfrac{a_{9}}{a_{6}}$

$\implies a_{9}= \dfrac{64 \times 64}{8} \implies \boxed{512}$

Let the $n$ th term of the geometric progression be $a_n=a_1r^{n-1}$ , where $r$ is the common ratio. Then,

$a_9 = a_1r^8 = a_1r^5 \cdot \frac {a_1r^5}{a_1r^2} = a_6 \cdot \frac {a_6}{a_3} = 64 \cdot \frac {64}8 = \boxed{512}$

Since you know that the $3$ rd term is $8$ and the $6$ th term is $64$ , the geometric progression is multiply by $\frac{8}{3}$ .

So:

$64 * 3($ $\frac{8}{3})$ $= 64 * 8 = \fbox {512}$

$a_n=a_m \times r^{n-m}$

$a_6=a_3 \times r^{6-3}$

$a_6=a_3 \times r^3$

$64=8 \times r^3$

$8=r^3$

$r=2$

$a_n=a_m \times r^{n-m}$

$a_9=a_3 \times r^{9-3}$

$a_9=8 \times 2^{6}$

$a_9=\boxed{512}$