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The equation provided is:

\[
K = \frac{a^3 \times \sqrt{b}}{\frac{1}{c^6} \times \frac{2}{d^5}}
\]

This expression can be simplified as follows:

1. **Simplify the denominator**:
   - The denominator has the form \(\frac{1}{c^6} \times \frac{2}{d^5}\).
   - Multiply these fractions: \(\frac{1 \times 2}{c^6 \times d^5} = \frac{2}{c^6d^5}\).

2. **Rewrite the entire expression**:
   - Now, the equation becomes:
   \[
   K = \frac{a^3 \times \sqrt{b}}{\frac{2}{c^6d^5}}
   \]
   - To simplify this, recall that dividing by a fraction is the same as multiplying by its reciprocal:
   \[
   K = a^3 \times \sqrt{b} \times \frac{c^6d^5}{2}
   \]

3. **Final simplified form**:
   - Combine the terms:
   \[
   K = \frac{a^3 \times c^6 \times d^5 \times \sqrt{b}}{2}
   \]

This is the simplified expression for \( K \).

Would you like to explore this equation further, or do you have any specific questions about the steps?

Here are five related questions:
1. How does the simplification process change if the exponents on \(c\) and \(d\) were different?
2. What happens to the expression if \( b = c^2 \)?
3. How can this expression be differentiated with respect to \( a \)?
4. What if \( a \), \( b \), \( c \), and \( d \) were specific numbers—how would you compute \( K \)?
5. How would this expression change if we used natural logarithms?

**Tip:** Always remember that multiplying by the reciprocal of a fraction is a useful technique for simplifying complex expressions.

K = \frac{a^3 \times \sqrt{b}}{\frac{1}{c^6} \times \frac{2}{d^5}}

Explore the simplification of the algebraic expression K = \frac{a^3 \times \sqrt{b}}{\frac{1}{c^6} \times \frac{2}{d^5}}. This problem delves into the core concepts of algebra, including the manipulation of exponents and radicals, suitable for high school students.

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