$\alpha=\max \left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\}$

$=\max \left\{2^{6 \sin 3 x} \cdot 2^{8 \cos 3 x}\right\}$

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

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

Now range of $6 \sin 3 x+8 \cos 3 x$

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

$\alpha=2^{10} \,\&\, \beta=2^{-10}$

$\text { So, } \alpha^{1 / 5}=2^{2}=4$

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

$\text { quadratic } 8 x^{2}+b x+c=0, c-b=$

$8 \times[\text { (product of roots })+(\text { sum of roots })]$

$=8 \times\left[4 \times \frac{1}{4}+4+\frac{1}{4}\right]=8 \times\left[\frac{21}{4}\right]=42$