$$ \sqrt{529}+\left(3^{}\right)^3-\left(6\right)^3 $$$$ =x^2 $$

Let's solve the equation step by step.

The given equation is:

\[
\sqrt{529} + 3^3 - 6^3 = x^2
\]

### Step 1: Calculate the square root
First, find the square root of 529:

\[
\sqrt{529} = 23
\]

### Step 2: Calculate the powers
Next, calculate $3^3$ and $6^3$:

\[
3^3 = 3 \times 3 \times 3 = 27
\]
\[
6^3 = 6 \times 6 \times 6 = 216
\]

### Step 3: Substitute and simplify
Now substitute the values back into the equation:

\[
23 + 27 - 216 = x^2
\]

Simplify the expression on the left side:

\[
50 - 216 = x^2
\]

\[
-166 = x^2
\]

### Step 4: Analyze the result
The equation $-166 = x^2$ implies that $x^2$ is negative. However, the square of a real number $x$ cannot be negative. Therefore, there is **no real solution** to this equation.

### Final Conclusion:
There is no real value of $x$ that satisfies this equation.

Would you like more details or have any other questions?

---

Here are some related questions:
1. Why can't the square of a real number be negative?
2. How do you find square roots of perfect squares?
3. What is the significance of powers in algebraic expressions?
4. Can you explain how to solve similar equations with positive results?
5. What are imaginary numbers, and how are they related to negative square roots?
6. How would this problem change if we dealt with complex numbers?
7. What would the solution be if the expression on the left side were positive?
8. Can you provide another example with different numbers?

**Tip:** When you encounter a square root of a negative number, consider the possibility of working with complex numbers.

Learn how to solve the equation involving square roots, exponentiation, and powers with a detailed step-by-step solution. This problem covers fundamental algebraic concepts, including square roots and powers. Suitable for students in Grades 7-9.

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