$$ \left(\sqrt7-\sqrt{-64}\right) $$($$ \sqrt7+\sqrt{-64}) $$

To simplify the given expression:

\[
\left( \sqrt{7} - \sqrt{-64} \right) \left( \sqrt{7} + \sqrt{-64} \right)
\]

we recognize that it has the form of a difference of squares, $ (a - b)(a + b) = a^2 - b^2 $, where $ a = \sqrt{7} $ and $ b = \sqrt{-64} $.

### Step 1: Simplify $ b = \sqrt{-64} $

We know that:

\[
\sqrt{-64} = \sqrt{64} \cdot \sqrt{-1} = 8i
\]

So the expression becomes:

\[
\left( \sqrt{7} - 8i \right) \left( \sqrt{7} + 8i \right)
\]

### Step 2: Apply the difference of squares formula

Using $ a = \sqrt{7} $ and $ b = 8i $, we apply the difference of squares formula:

\[
(\sqrt{7})^2 - (8i)^2
\]

### Step 3: Simplify the squares

1. $ (\sqrt{7})^2 = 7 $
2. $ (8i)^2 = 64i^2 = 64(-1) = -64 $

Now, the expression becomes:

\[
7 - (-64) = 7 + 64 = 71
\]

Thus, the simplified result is:

\[
\boxed{71}
\]

Would you like more details or have any questions? Here are some related questions:

1. What is the difference of squares in algebra?
2. How do you simplify square roots of negative numbers?
3. What are imaginary and complex numbers?
4. How do you multiply complex conjugates?
5. How does $i^2 = -1$ affect complex number calculations?

**Tip:** When simplifying expressions with complex numbers, always apply the rule $ i^2 = -1 $ to simplify terms involving imaginary numbers.

Learn how to simplify the expression $ \left( \sqrt{7} - \sqrt{-64} \right) \left( \sqrt{7} + \sqrt{-64} \right) $. Follow a step-by-step solution using the difference of squares formula and understanding of complex numbers. Suitable for high school students studying algebra and complex numbers.

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