Answer
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant.
Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{2^{(n+1)/3}}{2^{n/3}}=2^{1/3}$, thus it is a geometric sequence.
$a_1=\sqrt[3] 2$
$a_2=\sqrt[3] 4$
$a_3=2$
$a_4=2\sqrt[3] 2$
Work Step by Step
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant.
Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{2^{(n+1)/3}}{2^{n/3}}=2^{1/3}$, thus it is a geometric sequence.
$a_1=2^{1/3}=\sqrt[3] 2$
$a_2=2^{2/3}=\sqrt[3] 4$
$a_3=2^{3/3}=\sqrt[3] 8=2$
$a_4=2^{4/3}=\sqrt[3] 16=2\sqrt[3] 2$
