W1. Circle the first error made, if any, and provide the correct solution

a) (2 ^ 5 + 3 ^ 5 + 4 ^ 5) ^ 2

= 2 ^ 10 + 3 ^ 10 + 4 ^ 10 = 910 Step 1

Step 2

Let's go through the problem step by step to identify the first error.

### Original expression:
\[
(2^5 + 3^5 + 4^5)^2
\]

### Step 1 (according to the given solution):
\[
(2^5 + 3^5 + 4^5)^2 = 2^{10} + 3^{10} + 4^{10}
\]

### Issue:
This step is **incorrect**. You cannot distribute the exponentiation over addition like that. The rule for exponents says:
\[
(a + b + c)^2 \neq a^2 + b^2 + c^2
\]
Instead, you need to first calculate the values inside the parentheses and then square the result.

### Correct Solution:

#### Step 1: Compute each power inside the parentheses
\[
2^5 = 32, \quad 3^5 = 243, \quad 4^5 = 1024
\]
So,
\[
2^5 + 3^5 + 4^5 = 32 + 243 + 1024 = 1299
\]

#### Step 2: Square the result
\[
(1299)^2 = 1299 \times 1299 = 1,687,401
\]

Thus, the correct solution is:
\[
(2^5 + 3^5 + 4^5)^2 = 1,687,401
\]

Let me know if you need any further clarification or have any questions.

### Related Questions:
1. What is the difference between distributing multiplication and distributing exponents?
2. How can you verify the result of exponentiation using logarithms?
3. What are some properties of exponents that might help simplify more complex expressions?
4. How would you simplify \((x^2 + y^2 + z^2)^2\)?
5. What would happen if the powers were fractional instead of integers?

### Tip:
Always remember that exponentiation does not distribute over addition or subtraction unless the terms share a common factor.

Circle the first error made, if any, and provide the correct solution: a) (2^5 + 3^5 + 4^5)^2 = 2^10 + 3^10 + 4^10 = 910 Step 1

Learn how to correctly solve the expression (2^5 + 3^5 + 4^5)^2 by following the proper rules of exponentiation and order of operations. This detailed explanation identifies the first error and provides the correct solution, making it ideal for students in Grades 8-10.

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