Multiply. Simplify if possible.
\[
\frac{x^{2}-64}{x^{2}-7 x-8} \cdot \frac{x+1}{x}
\]
\[
\frac{x^{2}-64}{x^{2}-7 x-8} \cdot \frac{x+1}{x}=\square
\]
Answer
Final Answer: \(\boxed{\frac{x+8}{x}}\)
Step 1 ：Given the expression \(\frac{x^{2}-64}{x^{2}-7 x-8} \cdot \frac{x+1}{x}\)
Step 2 ：Factorize the numerator and denominator of the fractions. The expression \(x^{2}-64\) can be factorized as \((x-8)(x+8)\) and \(x^{2}-7x-8\) can be factorized as \((x-8)(x+1)\). So the expression becomes \(\frac{(x-8)(x+8)}{(x-8)(x+1)} \cdot \frac{x+1}{x}\)
Step 3 ：Cancel out the common factors in the numerator and denominator. The \((x-8)\) in the numerator and denominator cancel out, as does the \((x+1)\). So the expression simplifies to \(\frac{x+8}{x}\)
Step 4 ：Final Answer: \(\boxed{\frac{x+8}{x}}\)
