$$\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$

$$ = \max \{ {2^{6\sin 3x}}{.2^{8\cos 3x}}\} $$

$$ = max\{ {2^{6\sin 3x + 8\cos 3x}}\} $$

and $$\beta = \min \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} = \min \{ {2^{6\sin 3x + 8\cos 3x}}\} $$

Now range of $$6sin3x + 8cos3x$$

$$ = \left[ { - \sqrt {{6^2} + {8^2}} , + \sqrt {{6^2} + {8^2}} } \right] = [ - 10,10]$$

$$\alpha$$ = 2¹⁰ & $$\beta$$ = 2^$$-$$10

So, $$\alpha$$^1/5 = 2² = 4

$$\Rightarrow$$ $$\beta$$^1/5 = 2^$$-$$2 = 1/4

quadratic 8x² + bx + c = 0

$$ - {b \over 8} = {{17} \over 4}$$ $$ \Rightarrow $$ b = -34

$${c \over 8} = 1$$ $$ \Rightarrow $$ c = 8

$$ \therefore $$ c – b = 8 + 34 = 42