Okay, so I need to find the minimum value of the expression \( a^4 + b^8 \) given that \( ab^2 = 5 \), where \( a \) and \( b \) are positive real numbers. Let me start by understanding the problem. 

First, the constraint is \( ab^2 = 5 \). That means if I know one variable, I can find the other. For example, if I solve for \( a \), I get \( a = \frac{5}{b^2} \). Then I can substitute this into the expression \( a^4 + b^8 \) to make it a function of a single variable, either \( a \) or \( b \). That seems like a good approach. 

Let me try substituting \( a \) in terms of \( b \). So replacing \( a \) with \( \frac{5}{b^2} \) in the expression gives:

\( \left( \frac{5}{b^2} \right)^4 + b^8 = \frac{5^4}{b^8} + b^8 = \frac{625}{b^8} + b^8 \).

Now, the problem reduces to minimizing \( f(b) = \frac{625}{b^8} + b^8 \) where \( b > 0 \). Hmm, this looks like a calculus problem where I can take the derivative and set it equal to zero to find critical points. 

Alternatively, maybe I can use the AM-GM inequality here. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. But since there are two terms here, \( \frac{625}{b^8} \) and \( b^8 \), maybe I can apply AM-GM on these two terms. Let's see.

The AM-GM inequality for two terms says that:

\( \frac{\frac{625}{b^8} + b^8}{2} \geq \sqrt{\frac{625}{b^8} \cdot b^8} \).

Simplifying the right-hand side:

\( \sqrt{\frac{625}{b^8} \cdot b^8} = \sqrt{625} = 25 \).

Therefore, multiplying both sides by 2:

\( \frac{625}{b^8} + b^8 \geq 50 \).

So the minimum value of \( f(b) \) is 50? Wait, but hold on. The AM-GM equality holds when the two terms are equal. So, in this case, when \( \frac{625}{b^8} = b^8 \). Let me solve for \( b \) here.

Setting \( \frac{625}{b^8} = b^8 \), multiply both sides by \( b^8 \):

\( 625 = b^{16} \).

Taking the 16th root of both sides:

\( b = 625^{1/16} \).

But 625 is 5^4, so:

\( b = (5^4)^{1/16} = 5^{4/16} = 5^{1/4} = \sqrt[4]{5} \).

Then, substituting back into the expression for \( a \):

\( a = \frac{5}{b^2} = \frac{5}{(\sqrt[4]{5})^2} = \frac{5}{5^{1/2}} = 5^{1 - 1/2} = 5^{1/2} = \sqrt{5} \).

So, when \( a = \sqrt{5} \) and \( b = \sqrt[4]{5} \), the expression \( a^4 + b^8 \) equals 50. Therefore, the minimum value is 50. 

Wait, but let me verify this. Let me compute \( a^4 \) and \( b^8 \) with these values.

First, \( a = \sqrt{5} \), so \( a^4 = (\sqrt{5})^4 = (5^{1/2})^4 = 5^{2} = 25 \).

Then, \( b = \sqrt[4]{5} \), so \( b^8 = (\sqrt[4]{5})^8 = 5^{2} = 25 \). 

Adding them together: 25 + 25 = 50. Yep, that checks out. 

Alternatively, if I use calculus, let's confirm. Let's take the derivative of \( f(b) = \frac{625}{b^8} + b^8 \).

The derivative \( f'(b) = -625 \cdot 8 \cdot b^{-9} + 8b^7 = -\frac{5000}{b^9} + 8b^7 \).

Set derivative equal to zero:

\( -\frac{5000}{b^9} + 8b^7 = 0 \)

Multiply both sides by \( b^9 \):

\( -5000 + 8b^{16} = 0 \)

So, \( 8b^{16} = 5000 \)

Divide both sides by 8:

\( b^{16} = 625 \)

Taking the 16th root:

\( b = 625^{1/16} \)

Which is the same as before, since 625 is 5^4, so 625^{1/16} = 5^{4/16} = 5^{1/4} = \sqrt[4]{5}. 

Then, substituting back into \( a \), same result. 

Therefore, both methods confirm that the minimum value is 50. 

But let me think again. Is there a possibility that the minimum occurs at a boundary? For example, as \( b \) approaches 0 or infinity. 

If \( b \) approaches 0, then \( a = 5 / b^2 \) approaches infinity, so \( a^4 \) approaches infinity, and \( b^8 \) approaches 0. So the expression tends to infinity.

If \( b \) approaches infinity, then \( a = 5 / b^2 \) approaches 0, so \( a^4 \) approaches 0, and \( b^8 \) approaches infinity. Again, the expression tends to infinity.

Therefore, the function has a minimum somewhere in the middle, which we found using AM-GM or calculus. Therefore, the minimum value is indeed 50. 

Alternatively, maybe using substitution with exponents. Let me see. Since the constraint is \( ab^2 = 5 \), and we need to minimize \( a^4 + b^8 \). Let me express \( a \) and \( b \) in terms of exponentials. Let’s assume \( a = 5^{k} \) and \( b = 5^{m} \). Then the constraint becomes \( 5^{k} \cdot (5^{m})^2 = 5^{k + 2m} = 5^1 \), so \( k + 2m = 1 \). Then, the expression to minimize is \( (5^{k})^4 + (5^{m})^8 = 5^{4k} + 5^{8m} \). 

So we have to minimize \( 5^{4k} + 5^{8m} \) with \( k + 2m = 1 \). Let’s set \( k = 1 - 2m \), so the expression becomes \( 5^{4(1 - 2m)} + 5^{8m} = 5^{4 - 8m} + 5^{8m} \). Let’s denote \( x = 8m \), then the expression becomes \( 5^{4 - x} + 5^{x} \). Let’s set \( f(x) = 5^{4 - x} + 5^{x} \). To find the minimum, take derivative with respect to x:

\( f'(x) = -5^{4 - x} \ln 5 + 5^{x} \ln 5 \).

Set derivative to zero:

\( -5^{4 - x} \ln 5 + 5^{x} \ln 5 = 0 \)

Divide both sides by \( \ln 5 \):

\( -5^{4 - x} + 5^{x} = 0 \)

So, \( 5^{x} = 5^{4 - x} \)

Which implies \( x = 4 - x \), so \( 2x = 4 \), hence \( x = 2 \).

Therefore, \( 8m = 2 \) => \( m = 1/4 \), then \( k = 1 - 2*(1/4) = 1 - 1/2 = 1/2 \).

Therefore, \( a = 5^{1/2} = \sqrt{5} \), \( b = 5^{1/4} = \sqrt[4]{5} \). Which is the same as before. Then, the expression is \( 5^{4*(1/2)} + 5^{8*(1/4)} = 5^{2} + 5^{2} = 25 + 25 = 50 \). 

So, all methods lead to the same conclusion. Therefore, the minimum value is 50. 

I think that's thorough. I considered multiple approaches: substitution leading to a single variable function, AM-GM inequality, and calculus. All confirm that the minimum is 50. Also, checked the boundaries where the variables go to 0 or infinity, which confirm that the critical point found is indeed a minimum. Therefore, I can be confident the answer is 50.
Therefore, the minimum value of \( a^4 + b^8 \) is $\boxed{50}$.
</think>