Problem 145

(1) Since the sum of the coefficients in the expansion of $(3x - \frac{1}{3x})^n$ is 256, we can find the value of $n$ by considering the binomial expansion where $x=1$:

$$(3(1) - \frac{1}{3(1)})^n = 2^n = 256,$$

Solving the equation gives us:

$$n = \log_2{256} = 8.$$

Now, we know that the expansion will have $n + 1 = 9$ terms, and by using the symmetric property of binomial coefficients, the middle term when $n$ is an even number will have the greatest binomial coefficient. Therefore, the 5th term will have the maximum binomial coefficient since the terms are symmetric about the middle. The 5th term, $T_5$, is given by:

$$T_5 = \binom{8}{4}(3x)^{8-4} \left(-\frac{1}{3x}\right)^4 = \binom{8}{4} \cdot 3^4 \cdot x^4 \cdot \left(-\frac{1}{3}\right)^4 \cdot x^{-4}.$$

Simplifying, we find:

$$T_5 = \binom{8}{4} \cdot 3^4 \cdot \left(-\frac{1}{3}\right)^4 = \binom{8}{4} \cdot 3^4 \cdot 3^{-4} = \binom{8}{4} = 70.$$

The term with the maximum binomial coefficient (not accounting for the sign) is:

$$T_5 = 70 x^4 \cdot x^{-4} = \boxed{70}.$$

(2) To find the constant term, we examine the general term, $T_{r+1}$, of the binomial expansion:

$$T_{r+1} = \binom{8}{r} (3x)^{8-r} \left(-\frac{1}{3x}\right)^{r} = \binom{8}{r} \cdot 3^{8-r} \cdot (-1)^{r} \cdot x^{8-r} \cdot 3^{-r} \cdot x^{-r}.$$

Simplifying, we get:

$$T_{r+1} = \binom{8}{r} \cdot 3^{8-r} \cdot (-1)^{r} \cdot 3^{-r} \cdot x^{8-2r}.$$

For $T_{r+1}$ to be a constant term, the power of $x$ must be zero:

$$8 - 2r = 0,$$

Solving for $r$ yields:

$$r = 4.$$

Now, we can determine the constant term, which is $T_{4+1} = T_5$:

$$T_5 = \binom{8}{4} \cdot 3^{8-4} \cdot (-1)^4 \cdot 3^{-4} = \binom{8}{4} \cdot 3^{4} \cdot 3^{-4},$$

which simplifies to:

$$T_5 = \binom{8}{4} \cdot 1 = 70.$$

Therefore, the constant term in the expansion is:

$$\boxed{70}.$$