$$\begin{aligned} & 2 b=a+c \quad \text{.... (1)}\\ & b^2=(a+1)(c+3) \quad \text{.... (2)}\\ & \frac{a+b+c}{3}=8 \quad \text{.... (3)} \end{aligned}$$

$$\begin{aligned} \Rightarrow & \frac{3 b}{3}=8 \\ & b=8 \\ \Rightarrow \quad & a c+3 a+c+3=64 \end{aligned}$$

$$\begin{aligned} & 3 a+c+a c=61 \quad \text{... (4)}\\ & a+c=16 \\ & c=16-a \end{aligned}$$

from equation (4)

$$\begin{aligned} & 3 a+16-a+a(16-a)=61 \\ & \Rightarrow \quad(a-15)(a-3)=0 \\ & \quad a=15(a>10) \\ & \Rightarrow \quad a=15, b=8, c=1 \\ & \left((a \cdot b \cdot c)^{\frac{1}{3}}\right)^3=15 \times 8 \times 1=120 \end{aligned}$$