To simplify the expression

$\left(\frac{-eg^1 \cdot ah^{2/3}}{ge^{-5} \cdot h^{1/10} \cdot a^{2/3}}\right)^{-2}$

let's break it down step by step:

Step 1: Simplify the expression inside the parentheses.

First, we simplify the expression inside the parentheses:

$\frac{-e \cdot g \cdot a \cdot h^{2/3}}{g \cdot e^{-5} \cdot h^{1/10} \cdot a^{2/3}}$

Now, simplify each component:

1. Simplify the $e$-terms: $e^{1}$ in the numerator and $e^{-5}$ in the denominator.

$e^1 \cdot e^5 = e^{1 + 5} = e^6$

2. Simplify the $g$-terms: $g^1$ in the numerator and $g^1$ in the denominator.

$\frac{g^1}{g^1} = 1$

3. Simplify the $a$-terms: $a^1$ in the numerator and $a^{2/3}$ in the denominator.

$\frac{a^1}{a^{2/3}} = a^{1 - 2/3} = a^{1/3}$

4. Simplify the $h$-terms: $h^{2/3}$ in the numerator and $h^{1/10}$ in the denominator.

$h^{2/3 - 1/10} = h^{\frac{20}{30} - \frac{3}{30}} = h^{\frac{17}{30}}$

Now, combine all simplified components:

$\frac{-e^6 \cdot a^{1/3} \cdot h^{17/30}}{1} = -e^6 \cdot a^{1/3} \cdot h^{17/30}$

Step 2: Apply the exponent $-2$ to the simplified expression.

Now, raise the entire expression to the power of $-2$:

$\left(-e^6 \cdot a^{1/3} \cdot h^{17/30}\right)^{-2}$

Apply the exponent to each term individually:

1. $(-e^6)^{-2} = (-1)^{-2} \cdot e^{-12} = 1 \cdot e^{-12} = e^{-12}$
2. $(a^{1/3})^{-2} = a^{-2/3}$
3. $(h^{17/30})^{-2} = h^{-34/30} = h^{-17/15}$

So, the expression simplifies to:

$e^{-12} \cdot a^{-2/3} \cdot h^{-17/15}$

Final Answer:

$\frac{1}{e^{12} \cdot a^{2/3} \cdot h^{17/15}}$

Would you like further details or have any questions?

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Here are 5 related questions:

1. How would the expression change if the power was positive instead of negative?
2. What if the base $a$ was a variable with a negative exponent originally?
3. How do you simplify expressions with fractional exponents?
4. How do you handle expressions with roots and negative exponents together?
5. What is the physical interpretation of negative exponents in real-world applications?

Tip: When simplifying expressions with multiple terms in the numerator and denominator, always handle each variable separately to avoid mistakes.