The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27/19. Then the common ratio of this series is:

Let the first term of the infinite geometric series be 'a' and the common ratio be 'r'.
Since the sum of the series is 3, we have:

a/(1-r) = 3 (1)

The sum of the cubes of the terms is given by:

a³/(1-r³) = 27/19 (2)

From equation (1), we have a = 3(1-r).
Substituting this into equation (2), we get:

[3(1-r)]³/(1-r³) = 27/19

27(1-r)³/(1-r³)=27/19

19(1-r)³ = 1-r³

19(1 - 3r + 3r² - r³) = 1 - r³

19 - 57r + 57r² - 19r³ = 1 - r³

18r³ - 57r² + 57r - 18 = 0

Dividing by 3:

6r³ - 19r² + 19r - 6 = 0

By observation, r = 2/3 is a root.
We can factor the cubic equation as follows:

(3r-2)(2r²-5r+3) = 0

(3r-2)(2r-3)(r-1) = 0

The possible values of r are 2/3, 3/2, and 1. Since |r| < 1 for an infinite geometric series to converge and the terms are positive, we must have r = 2/3.
Therefore, the common ratio is 2/3 or 23.