The expression given is $(8x^4 + 1)^2$. To expand this, we will use the algebraic identity for the square of a binomial:

$(a + b)^2 = a^2 + 2ab + b^2$

In this case, $a = 8x^4$ and $b = 1$.

Let's expand it step by step:

$(8x^4 + 1)^2 = (8x^4)^2 + 2 \cdot (8x^4) \cdot 1 + 1^2$

1. First, calculate $(8x^4)^2$: $(8x^4)^2 = 64x^8$

2. Next, calculate $2 \cdot (8x^4) \cdot 1$: $2 \cdot 8x^4 \cdot 1 = 16x^4$

3. Finally, calculate $1^2$: $1^2 = 1$

Now, combine all the terms:

$(8x^4 + 1)^2 = 64x^8 + 16x^4 + 1$

So the expanded form of the expression is:

$64x^8 + 16x^4 + 1$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How would you expand $(2x^2 + 3)^2$?
2. What is the general formula for expanding $(ax^n + b)^2$?
3. How does this expansion relate to the binomial theorem?
4. Can you factorize $64x^8 + 16x^4 + 1$ further?
5. What happens if you multiply $(8x^4 + 1)$ by another polynomial?

Tip: When expanding binomials, ensure that you correctly apply the identity $(a + b)^2 = a^2 + 2ab + b^2$, as this helps in avoiding common mistakes.