If \( a = b = c \) and \( a^2 + b^2 + c^2 = 996 \), we can find \( a + b + c \) by following these steps:

1. Since \( a = b = c \), we can express the sum of squares as:
   \[
   a^2 + b^2 + c^2 = 3a^2
   \]
2. We know that \( 3a^2 = 996 \).

3. Solving for \( a^2 \), we divide both sides by 3:
   \[
   a^2 = \frac{996}{3} = 332
   \]

4. Now, take the square root of both sides to find \( a \):
   \[
   a = \sqrt{332}
   \]

   The square root of 332 is approximately \( a \approx 18.22 \).

5. Finally, since \( a = b = c \), the sum \( a + b + c \) is:
   \[
   a + b + c = 3a = 3 \times 18.22 \approx 54.66
   \]

So, \( a + b + c \approx 54.66 \).

Would you like to explore this further or have any specific questions?

Here are five related questions:
1. How does the result change if \( a \), \( b \), and \( c \) are different?
2. What if \( a^2 + b^2 + c^2 = 1000 \) instead of 996?
3. How do you calculate \( a \) exactly if the square root isn't a whole number?
4. Can you solve for \( a + b + c \) if you know \( a^2 + b^2 + c^2 \) but only two of the values?
5. How does this solution relate to the concept of Pythagorean triples?

**Tip:** When dealing with equations involving squares, it's often helpful to simplify by assuming equal values, as it can make the calculation much easier.

Learn how to calculate the sum a + b + c given that a^2 + b^2 + c^2 = 996 and a = b = c. Follow a step-by-step solution using the Pythagorean theorem and algebraic simplification. Suitable for Grades 7-9 students.

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