Use the properties of exponents to simplify ((-8 m^5 n^8) / (2 m^7 n^-3))^-4. Simplify inside the parentheses first.

Step-by-step solution

1. 1

   Simplify inside before touching the outer exponent. Raising a messy quotient to $-4$ invites mistakes; reducing the inside to a single monomial first makes the last step mechanical. Inside, handle the coefficient and each variable separately.

2. 2

   Divide the coefficients and subtract the exponents. For like bases, $\dfrac{a^p}{a^q} = a^{p-q}$:

   $\frac{-8}{2} = -4, \qquad m^{5-7} = m^{-2}, \qquad n^{8-(-3)} = n^{11}$

   The $n$ exponent is the trap: $8-(-3) = 11$, not $5$. So the bracket becomes

   $\left(-4m^{-2}n^{11}\right)^{-4}$

3. 3

   Distribute the outer $-4$ to every factor. Using $(abc)^k = a^k b^k c^k$ and $(a^p)^k = a^{pk}$:

   $(-4)^{-4} \cdot m^{(-2)(-4)} \cdot n^{(11)(-4)} = (-4)^{-4}\, m^{8}\, n^{-44}$

4. 4

   Evaluate the numeric factor. An even negative exponent kills the minus sign:

   $(-4)^{-4} = \frac{1}{(-4)^4} = \frac{1}{256}$

   Writing $-\tfrac{1}{256}$ is the classic error — the exponent $4$ is even, so the result is positive.

5. 5

   Clear the remaining negative exponent. Move $n^{-44}$ into the denominator:

   $\frac{m^8}{256\,n^{44}}$

6. 6

   Check with numbers. Put $m = 2$, $n = 3$. Inside: $\dfrac{-8 \cdot 32 \cdot 6561}{2 \cdot 128 \cdot \frac{1}{27}} = -4 \cdot 2^{-2} \cdot 3^{11} = -177147$. Then $(-177147)^{-4} = 1/984{,}770{,}902{,}183{,}611{,}232{,}881$. The answer gives $\dfrac{2^8}{256 \cdot 3^{44}} = \dfrac{1}{3^{44}}$, and $3^{44} = 984{,}770{,}902{,}183{,}611{,}232{,}881$ ✓ — the two agree exactly.