2021 AMC 10A Problem 10

1. Try a small version first. The first factor is $2 + 3 = 5$, and $3^2 - 2^2 = 9 - 4 = 5$ too.
2. Add the next factor: $(2 + 3)(2^2 + 3^2) = 5 \times 13 = 65$, and $3^4 - 2^4 = 81 - 16 = 65$. Again they match. It looks like the product is always "$3$ to some power minus $2$ to the same power".
3. Here is why. The difference of squares rule says $(a - b)(a + b) = a^2 - b^2$. Multiply the whole product by $(3 - 2)$, which equals $1$, so nothing changes. Then combine from the left:
$$(3 - 2)(3 + 2) = 3^2 - 2^2,$$
$$(3^2 - 2^2)(3^2 + 2^2) = 3^4 - 2^4,$$
$$(3^4 - 2^4)(3^4 + 2^4) = 3^8 - 2^8.$$
4. Each factor doubles the exponent: $2, 4, 8, 16, 32, 64, 128$. The last factor is $(2^{64} + 3^{64})$, and combining with it gives
$$(3^{64} - 2^{64})(3^{64} + 2^{64}) = 3^{128} - 2^{128}.$$

The answer is $\boxed{\textbf{(C)}\ 3^{128}-2^{128}}$.

A product like $(a + b)(a^2 + b^2)(a^4 + b^4)\cdots$ collapses when you multiply by $(a - b)$, because each step is a difference of squares. Here $a - b = 3 - 2 = 1$, so multiplying by it is free. There are seven factors, and seven doublings take the exponent from $1$ to $2^7 = 128$, not $127$.