The problem asks us to factor the expression $192s^3 + 3$ using the sum of cubes formula and to factor out any common factors.

1. Problem Description

The problem asks us to factor the expression $192s^3 + 3$ using the sum of cubes formula and to factor out any common factors.

2. Solution Steps

First, we can factor out the common factor 3 from both terms:

$192s^3 + 3 = 3(64s^3 + 1)$

Now, we can rewrite $64s^3 + 1$ as $(4s)^3 + (1)^3$. We can use the sum of cubes formula:

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

In this case, $a = 4s$ and $b = 1$. Applying the formula:

$(4s)^3 + 1^3 = (4s + 1)((4s)^2 - (4s)(1) + (1)^2)$

$= (4s + 1)(16s^2 - 4s + 1)$

So, $64s^3 + 1 = (4s + 1)(16s^2 - 4s + 1)$.

Finally, we substitute this back into the expression with the factored out 3:

$3(64s^3 + 1) = 3(4s + 1)(16s^2 - 4s + 1)$

3. Final Answer

$3(4s+1)(16s^2-4s+1)$