If $$\log_3(x^2 - 1), \log_3(2x^2 + 1)$$ and $$\log_3(6x^2 + 3)$$ are the first three terms of an arithmetic progression, then the sum of the next three terms of the progression is

$$\log_3(x^2 - 1), \log_3(2x^2 + 1)$$ and $$\log_3(6x^2 + 3)$$ are the first three terms of an arithmetic progression.

=> $$\log_3(2x^2 + 1)-\log_3(x^2 - 1)=\log_3(6x^2 + 3)-\log_3(2x^2 + 1)$$

=> $$\log_3\left[\dfrac{2x^2+1}{x^2-1}\right]=\log_3\left[\dfrac{6x^2+3}{2x^2+1}\right]$$

=> $$\log_3\left[\dfrac{2x^2+1}{x^2-1}\right]=\log_33$$

=> $$\dfrac{2x^2+1}{x^2-1}=3$$

=> $$2x^2+1=3x^2-3$$

=> $$x^2=4$$

Thus, the first three terms of the AP will be $$\log_3(x^2 - 1)=1$$, $$\log_3(2x^2 + 1)=2$$, $$\log_3(6x^2 + 3)=3$$.

Therefore, the next three terms of the AP will be 4, 5, and 6, and their sum will be = 4 + 5 + 6 = 15.