Let $$\alpha = \max_{x \in R}\{8^{2\sin 3x} \cdot 4^{4\cos 3x}\}$$ and $$\beta = \min_{x \in R}\{8^{2\sin 3x} \cdot 4^{4\cos 3x}\}$$. If $$8x^2 + bx + c = 0$$ is a quadratic equation whose roots are $$\alpha^{1/5}$$ and $$\beta^{1/5}$$, then the value of $$c - b$$ is equal to:

First, let us rewrite the given expression in a single base so that the maximisation and minimisation become easier.

We have $$8^{2\sin 3x}\cdot 4^{4\cos 3x}.$$

Recall that $$8 = 2^3 \quad\text{and}\quad 4 = 2^2.$$ Substituting these powers of two, we get

$$\bigl(2^3\bigr)^{2\sin 3x}\cdot\bigl(2^2\bigr)^{4\cos 3x} = 2^{3\cdot 2\sin 3x}\cdot 2^{2\cdot 4\cos 3x} = 2^{6\sin 3x}\cdot 2^{8\cos 3x}.$$

Whenever we multiply two numbers with the same base, we add the exponents, so

$$2^{6\sin 3x}\cdot 2^{8\cos 3x}=2^{\,6\sin 3x+8\cos 3x}.$$ Hence, maximising or minimising the original expression is the same as maximising or minimising the exponent

$$f(t)=6\sin t+8\cos t,$$

where we have set $$t = 3x,$$ because $$\sin 3x$$ and $$\cos 3x$$ both depend only on $$3x.$$

Now, for any expression of the form $$A\sin t + B\cos t,$$ its maximum value is $$\sqrt{A^{2}+B^{2}}$$ and its minimum value is $$-\sqrt{A^{2}+B^{2}}.$$ This comes directly from interpreting the linear combination as a shifted sine wave with amplitude $$\sqrt{A^{2}+B^{2}}.$$ Stating the result clearly,

$$\max(A\sin t+B\cos t)=\sqrt{A^2+B^2},\qquad \min(A\sin t+B\cos t)=-\sqrt{A^2+B^2}.$$

Here, $$A=6$$ and $$B=8,$$ so

$$\sqrt{A^{2}+B^{2}}=\sqrt{6^{2}+8^{2}} =\sqrt{36+64} =\sqrt{100}=10.$$

Therefore

$$\max_{t\in\mathbb R} f(t)=10,\qquad \min_{t\in\mathbb R} f(t)=-10.$$

Remembering that the original quantity is $$2^{f(t)},$$ we get

$$\alpha = 2^{10}=1024,\qquad \beta = 2^{-10}=\dfrac1{1024}.$$

The problem says that $$\alpha^{1/5}$$ and $$\beta^{1/5}$$ are the roots of the quadratic $$8x^{2}+bx+c=0.$$ Let us denote the two roots by

$$r_{1}=\alpha^{1/5},\qquad r_{2}=\beta^{1/5}.$$

We now compute these roots explicitly. Since $$\alpha=1024=2^{10},$$

$$r_{1}=\bigl(2^{10}\bigr)^{1/5}=2^{10/5}=2^{2}=4.$$

Similarly, $$\beta=2^{-10},$$ so

$$r_{2}=\bigl(2^{-10}\bigr)^{1/5}=2^{-10/5}=2^{-2}=\dfrac14.$$

Next, recall the standard relations between the coefficients of a quadratic and its roots. For a quadratic $$ax^{2}+bx+c=0$$ with roots $$r_{1},\,r_{2},$$ we have

$$r_{1}+r_{2}=-\dfrac{b}{a},\qquad r_{1}r_{2}=\dfrac{c}{a}.$$

In our particular equation, $$a=8,$$ so

$$r_{1}+r_{2}=-\dfrac{b}{8},\qquad r_{1}r_{2}=\dfrac{c}{8}.$$

Substituting the numerical values we have found:

Sum of roots: $$r_{1}+r_{2}=4+\dfrac14=\dfrac{16}{4}+\dfrac14=\dfrac{17}{4}.$$

Thus $$-\dfrac{b}{8}=\dfrac{17}{4}\quad\Longrightarrow\quad b=-8\cdot\dfrac{17}{4}=-34.$$

Product of roots: $$r_{1}r_{2}=4\cdot\dfrac14=1.$$

Thus $$\dfrac{c}{8}=1\quad\Longrightarrow\quad c=8.$$

The question finally asks for $$c-b.$$ We compute

$$c-b = 8 - (-34)=8+34=42.$$

Hence, the correct answer is Option A.