https://file.aiquickdraw.com/mathbot/file/1726502328497c4kaiumf.jpg

The mathematical expression in the image is:

\[
\left( \frac{7a^2 b^3 c^0}{3a^3 b^4 c^3} \right)^{-4}
\]

### Simplifying Step-by-Step:

1. **Simplify Inside the Parentheses**:
   - \( c^0 = 1 \) (anything raised to the power of 0 is 1).
   - So, we simplify:
     \[
     \frac{7a^2 b^3}{3a^3 b^4 c^3}
     \]
   
2. **Divide the Like Terms**:
   - For \( a \)'s: \(\frac{a^2}{a^3} = a^{2-3} = a^{-1} \).
   - For \( b \)'s: \(\frac{b^3}{b^4} = b^{3-4} = b^{-1} \).
   - For \( c \)'s: \(\frac{1}{c^3} = c^{-3} \).

   So the expression simplifies to:
   \[
   \frac{7}{3} a^{-1} b^{-1} c^{-3}
   \]

3. **Apply the Exponent Outside the Parentheses**:
   The entire expression is raised to the power of \(-4\):
   \[
   \left( \frac{7}{3} a^{-1} b^{-1} c^{-3} \right)^{-4}
   \]
   
4. **Distribute the Exponent**:
   - \( \left( \frac{7}{3} \right)^{-4} = \frac{3^4}{7^4} = \frac{81}{2401} \).
   - \( (a^{-1})^{-4} = a^4 \).
   - \( (b^{-1})^{-4} = b^4 \).
   - \( (c^{-3})^{-4} = c^{12} \).

   So, the expression becomes:
   \[
   \frac{81}{2401} a^4 b^4 c^{12}
   \]

### Final Answer:
\[
\boxed{\frac{81 a^4 b^4 c^{12}}{2401}}
\]

Let me know if you would like further details or have any questions!

### Related Questions:
1. How do you simplify negative exponents when they appear in a denominator?
2. What is the rule for dividing terms with the same base but different exponents?
3. How do you handle exponents when raising fractions to a power?
4. Can you explain why anything raised to the power of zero equals one?
5. How does raising a term with a negative exponent to a power affect its sign?

### Tip:
When simplifying fractions with exponents, remember to subtract the exponents of like bases (numerator minus denominator).

Simplify the expression (7a^2 b^3 c^0 / 3a^3 b^4 c^3)^-4.

Learn how to simplify the expression (7a^2 b^3 c^0 / 3a^3 b^4 c^3)^-4 by using exponent rules, including handling zero and negative exponents. This detailed step-by-step solution is ideal for high school algebra students.

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