To simplify the expression \((3x + 1)^2(3x + 1)^2\), we can first notice that we are multiplying the same term.

We can combine the powers as follows:

\[
(3x + 1)^2(3x + 1)^2 = (3x + 1)^{2 + 2} = (3x + 1)^4
\]

Now, if you want to expand \((3x + 1)^4\), we can use the binomial theorem or expand it stepwise. Here, I'll use the binomial theorem for expansion:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

For \((3x + 1)^4\):

- \(a = 3x\)
- \(b = 1\)
- \(n = 4\)

The expansion is:

\[
(3x + 1)^4 = \sum_{k=0}^{4} \binom{4}{k} (3x)^{4-k} (1)^k
\]

Calculating each term:

- For \(k = 0\): \(\binom{4}{0}(3x)^4(1)^0 = 1 \cdot 81x^4 \cdot 1 = 81x^4\)
- For \(k = 1\): \(\binom{4}{1}(3x)^3(1)^1 = 4 \cdot 27x^3 \cdot 1 = 108x^3\)
- For \(k = 2\): \(\binom{4}{2}(3x)^2(1)^2 = 6 \cdot 9x^2 \cdot 1 = 54x^2\)
- For \(k = 3\): \(\binom{4}{3}(3x)^1(1)^3 = 4 \cdot 3x \cdot 1 = 12x\)
- For \(k = 4\): \(\binom{4}{4}(3x)^0(1)^4 = 1 \cdot 1 \cdot 1 = 1\)

Now, combining all these terms together:

\[
(3x + 1)^4 = 81x^4 + 108x^3 + 54x^2 + 12x + 1
\]

Thus, the simplified form of the original expression \((3x + 1)^2(3x + 1)^2\) is:

\[
(3x + 1)^4 = 81x^4 + 108x^3 + 54x^2 + 12x + 1
\]