Question

$a + b + c = 3a \Rightarrow b + c = 2a$

$$\begin{matrix} abc = c^{3} \Rightarrow ab = c^{2}\ \text{~or~}c = 0 \Rightarrow a = 0,b = 0\text{~or~}a = 3,b = 6 \\ ab + bc + ca = 3b \\ (a,b,c) = (0,0,0),(3,6,0) \end{matrix}$$

OR

$$\begin{matrix} c^{2} + bc + ca = 3b = c(3a)\ \Rightarrow \ b = ac \\ b + c = 2a,b = ac,c^{2} = ab \\ b + c^{2} = a(b + c) = 2a^{2} \end{matrix}$$

$$\begin{matrix} \therefore & a = c = b\text{~or~}c = - 2a \\ \Rightarrow & b = 4a \\ & 4a = - 2a^{2}\ \Rightarrow \ a = 0,a = - 2. \end{matrix}$$

$$\begin{matrix} (a,\text{ }b,c) = ( - 2, - 8,4) \\ x^{3} + 6x^{2} - 24x - 64 = (x + 2)(x + 8)(x - 4) \end{matrix}$$