Solve the following inequality.
\[
(x-7)^{2}(x+8)> 0
\]

Step 1 ：The inequality is a product of two factors: \((x-7)^2\) and \((x+8)\). The inequality is greater than zero, which means we are looking for the x-values where the expression is positive.

Step 2 ：The expression \((x-7)^2\) is always non-negative because it is a square of a real number. It is zero when \(x=7\) and positive otherwise.

Step 3 ：The expression \((x+8)\) is negative when \(x<-8\), zero when \(x=-8\), and positive when \(x>-8\).

Step 4 ：So, we need to find the intervals of \(x\) where the product of these two expressions is positive.

Step 5 ：The solution indicates that the inequality is satisfied when \(x > -8\) and \(x \neq 7\). This means that the product of the two expressions is positive when \(x\) is greater than -8 but not equal to 7.

Step 6 ：Final Answer: The solution to the inequality \((x-7)^{2}(x+8)>0\) is \(x \in (-8, 7) \cup (7, \infty)\), or in interval notation, \((-8, 7) \cup (7, \infty)\).