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The equation provided is:

\[
T = \sqrt[7x]{\frac{(81)^{x + \frac{1}{4}}}{3^{x+1} \cdot \sqrt[3]{3^{-3}}}}
\]

Let me simplify and explain it step by step.

### Step-by-step simplification:
1. **Simplifying \( 81^{x + \frac{1}{4}} \)**:  
   We know that \( 81 = 3^4 \), so:
   \[
   81^{x + \frac{1}{4}} = (3^4)^{x + \frac{1}{4}} = 3^{4(x + \frac{1}{4})} = 3^{4x + 1}
   \]

2. **Simplifying \( \sqrt[3]{3^{-3}} \)**:  
   This simplifies to:
   \[
   \sqrt[3]{3^{-3}} = (3^{-3})^{\frac{1}{3}} = 3^{-1}
   \]

3. **Substituting these into the original equation**:
   Now we substitute the simplified terms into the equation:
   \[
   T = \sqrt[7x]{\frac{3^{4x+1}}{3^{x+1} \cdot 3^{-1}}}
   \]

4. **Combining powers of 3 in the denominator**:  
   We combine the powers of 3 in the denominator:
   \[
   3^{x+1} \cdot 3^{-1} = 3^{(x+1) - 1} = 3^x
   \]

5. **Final simplification**:
   Now the equation becomes:
   \[
   T = \sqrt[7x]{\frac{3^{4x+1}}{3^x}} = \sqrt[7x]{3^{(4x+1) - x}} = \sqrt[7x]{3^{3x+1}}
   \]

6. **Simplifying the root**:
   Taking the 7x-th root of \( 3^{3x+1} \):
   \[
   T = 3^{\frac{3x+1}{7x}}
   \]

Thus, the simplified form of the equation is:
\[
T = 3^{\frac{3x+1}{7x}}
\]

Would you like further details or have any questions?

Here are five related questions you might explore:
1. How does the property \( (a^m)^n = a^{mn} \) help in simplifying exponential expressions?
2. Can you explain why the cube root of \( 3^{-3} \) is \( 3^{-1} \)?
3. How does the general property of fractional exponents \( \sqrt[n]{a^m} = a^{m/n} \) apply here?
4. What happens to the equation if you change the base from 3 to another number, say 5?
5. How would you approach this problem if you had logarithms involved?

**Tip**: Understanding how to manipulate exponents is key when simplifying complex algebraic expressions.

Simplify the given equation: T = sqrt[7x]{(81)^(x + 1/4) / (3^(x+1) * sqrt[3]{3^(-3)})}

Learn how to simplify the exponential equation T = sqrt[7x]{(81)^(x + 1/4) / (3^(x+1) * sqrt[3]{3^(-3)})} by applying exponent and root properties. This detailed solution is suitable for students in Grades 10-12 studying algebra and exponentiation.

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