Exponent and Cube Root Divisibility Condition

Simplify the expression under the cube root: $(3)^a (3 \cdot 7)^{(3a-b)} (7^2)^{(2b+c)} = 3^{a + 3a - b} \cdot 7^{3a - b + 4b + 2c} = 3^{4a - b} \cdot 7^{3a + 3b + 2c}$.

For the cube root $\sqrt[3]{3^{4a-b} \cdot 7^{3a+3b+2c}}$ to be an integer, the exponents of prime factors 3 and 7 must be multiples of 3:

1. $4a - b = 3a + (a - b)$ is a multiple of $3 \implies a - b$ must be a multiple of 3.
2. $3a + 3b + 2c = 3(a + b) + 2c$ is a multiple of $3 \implies 2c$ must be a multiple of $3 \implies c$ must be a multiple of 3.

Thus, $(a - b)$ and $c$ being divisible by 3 is both necessary and sufficient.