To expand the binomial $(3x^2 + 1)^3$ using the Binomial Theorem, we'll apply the formula:

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

Here, $a = 3x^2$, $b = 1$, and $n = 3$.

Let's calculate each term:

1. Term 1: $k = 0$
   $\binom{3}{0} (3x^2)^{3} (1)^{0} = 1 \times (3x^2)^3 = 27x^6$

2. Term 2: $k = 1$
   $\binom{3}{1} (3x^2)^{2} (1)^1 = 3 \times (3x^2)^2 = 27x^4$

3. Term 3: $k = 2$
   $\binom{3}{2} (3x^2)^{1} (1)^2 = 3 \times (3x^2) = 9x^2$

4. Term 4: $k = 3$
   $\binom{3}{3} (3x^2)^{0} (1)^3 = 1 \times 1 = 1$

Now, we combine all these terms:

$(3x^2 + 1)^3 = 27x^6 + 27x^4 + 9x^2 + 1$

Do you want more details or have any questions?

Here are some related questions:

1. How do you expand a binomial using the Binomial Theorem?
2. What are binomial coefficients, and how are they calculated?
3. How would you expand $(2x + 3)^4$ using the Binomial Theorem?
4. What is Pascal’s Triangle, and how does it relate to binomial expansions?
5. How do you find the middle term in a binomial expansion?

Tip: Remember that the sum of the exponents in each term of a binomial expansion equals the power $n$ of the binomial.