To solve the given problem systematically, we will address each part in sequence:

1. Solve $\frac{a^2}{b} - \frac{b^2}{a} = 3(a - b)$

Step 1: Simplify the equation

The equation can be rewritten with a common denominator: $\frac{a^3 - b^3}{ab} = 3(a - b)$

Step 2: Factorize the numerator

The expression $a^3 - b^3$ can be factored as: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ So the equation becomes: $\frac{(a - b)(a^2 + ab + b^2)}{ab} = 3(a - b)$

Step 3: Simplify

If $a \neq b$, cancel $a - b$: $\frac{a^2 + ab + b^2}{ab} = 3$

Multiply through by $ab$: $a^2 + ab + b^2 = 3ab$

Rearrange terms: $a^2 - 2ab + b^2 = 0$

Step 4: Recognize a perfect square

$(a - b)^2 = 0$

So, $a = b$.

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2. Solve $\frac{4(a^8 + b^8)}{(a^2b^2)^2}$

Step 1: Substitute $a = b$

If $a = b$, then $a^8 + b^8 = 2a^8$, and $(a^2b^2)^2 = (a^4)^2 = a^8$. Substituting into the expression: $\frac{4(a^8 + b^8)}{(a^2b^2)^2} = \frac{4(2a^8)}{a^8} = \frac{8a^8}{a^8} = 8$

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Final Answer:

$\boxed{8}$

Let me know if you'd like any part of this solution elaborated further!

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Related Questions:

1. What happens if $a \neq b$ in the first equation?
2. How can we generalize the simplifications for $\frac{a^2}{b} - \frac{b^2}{a} = 3(a - b)$?
3. Explore other factorizations for $a^3 - b^3$.
4. How does symmetry in $a = b$ simplify calculations for higher powers?
5. How would the solution differ if $4(a^8 + b^8)$ included additional terms?

Tip:

Always look for symmetry in equations—it can often simplify the algebra significantly!