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Step-by-Step Solution
Step 1: Express the expression inside the maximum and minimum
We have
$$
\alpha = \max \limits_{x \in \mathbb{R}} \Bigl\{ 8^{2 \sin 3x} \cdot 4^{4 \cos 3x} \Bigr\}
\quad\text{and}\quad
\beta = \min \limits_{x \in \mathbb{R}} \Bigl\{ 8^{2 \sin 3x} \cdot 4^{4 \cos 3x} \Bigr\}.
$$
Notice that
$$
8^{2 \sin 3x} = (2^3)^{2 \sin 3x} = 2^{6 \sin 3x}
\quad\text{and}\quad
4^{4 \cos 3x} = (2^2)^{4 \cos 3x} = 2^{8 \cos 3x}.
$$
Therefore,
$$
8^{2 \sin 3x} \cdot 4^{4 \cos 3x}
= 2^{6 \sin 3x} \cdot 2^{8 \cos 3x}
= 2^{6 \sin 3x + 8 \cos 3x}.
$$
Hence,
$$
\alpha = \max \limits_{x \in \mathbb{R}} \{\,2^{6 \sin 3x + 8 \cos 3x}\}
\quad\text{and}\quad
\beta = \min \limits_{x \in \mathbb{R}} \{\,2^{6 \sin 3x + 8 \cos 3x}\}.
$$
Step 2: Determine the range of $6\,\sin 3x + 8\,\cos 3x$
For any real $x,$ the expression $a \sin \theta + b \cos \theta$ takes values in the interval
$[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}].$
Here, $a = 6$ and $b = 8.$ Thus,
$$
6 \sin 3x + 8 \cos 3x \in
\bigl[-\sqrt{6^2 + 8^2}, \sqrt{6^2 + 8^2}\bigr]
= [-10,\,10].
$$
Step 3: Determine the maximum and minimum values of the original expression
Since $6 \sin 3x + 8 \cos 3x$ ranges from $-10$ to $10,$
$$
2^{6 \sin 3x + 8 \cos 3x}
\quad\text{ranges from}\quad
2^{-10} \quad\text{to}\quad 2^{10}.
$$
Hence
$$
\alpha = 2^{10}
\quad\text{and}\quad
\beta = 2^{-10}.
$$
Step 4: Find the values of $\alpha^{1/5}$ and $\beta^{1/5}$
We are told the roots of the quadratic equation
$$
8x^2 + bx + c = 0
$$
are $\alpha^{1/5}$ and $\beta^{1/5}.$ Substituting the values of $\alpha$ and $\beta$:
$$
\alpha^{1/5} = \bigl(2^{10}\bigr)^{1/5} = 2^{10/5} = 2^2 = 4,
$$
$$
\beta^{1/5} = \bigl(2^{-10}\bigr)^{1/5} = 2^{-10/5} = 2^{-2} = \tfrac{1}{4}.
$$
Step 5: Sum and product of the roots in the quadratic equation
For a quadratic equation
$$
8x^2 + bx + c = 0,
$$
the sum of the roots is
$$
\alpha^{1/5} + \beta^{1/5} = 4 + \tfrac{1}{4} = \tfrac{17}{4}.
$$
According to Vieta's formulas:
$$
\alpha^{1/5} + \beta^{1/5} = -\frac{b}{8}.
$$
Hence
$$
-\frac{b}{8} = \tfrac{17}{4}
\quad\Longrightarrow\quad
\frac{b}{8} = -\tfrac{17}{4}
\quad\Longrightarrow\quad
b = -34.
$$
The product of the roots is
$$
\alpha^{1/5} \cdot \beta^{1/5} = 4 \times \frac{1}{4} = 1.
$$
Again, by Vieta's formulas:
$$
\alpha^{1/5} \cdot \beta^{1/5} = \frac{c}{8}.
$$
Therefore,
$$
\frac{c}{8} = 1
\quad\Longrightarrow\quad
c = 8.
$$
Step 6: Compute $c - b$
Substituting the values of $b$ and $c,$ we get
$$
b = -34 \quad\text{and}\quad c = 8.
$$
Therefore,
$$
c - b = 8 - (-34) = 8 + 34 = 42.
$$
Final Answer
The value of $c - b$ is 42.
