Algebra 2-Factoring sum of cubes by grouping Factor the sum  of cubes: $81x^3+192$
After finding the prime factorization of both numbers I found that $81$ is $3^4$
and $192$ is $2^6 \cdot 3$. 
The problem is I tried grouping and found $3$ is the LCM so it would outside in parenthesis. The formula for the sum of cubes is $(a+b) (a^2-ab+b^2)=a^3+b^3$
So I tried writing it this way and got:$(3x+2) (9x^2-6x+4)$ and then I realized I was wrong because $b$ from the formula (sum of cubes) was wrong. So how do I find $b$? and Why did I get $2$ for $b$ in the formula?

$$81x^3+192 = 3 (27 x^3 + 64) = 3 ((3x)^3+4^3) \\= 3 (3x + 4) ((3x)^2 - 3x\cdot 4 + 4^2)  = 3 (3x+4)(9x^2-12x+16)$$
Since $12^2-4\cdot9\cdot16$ does not have a nice square root further factorization is not possible.
