Let α = max _ x ∈ R \ 8 ^ 2 3x · 4 ^ 4 3x \ and β = min _ x ∈ R \ 8...

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

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

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

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

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

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

$\alpha = {2}^{10} \text{ and } \beta = {2}^{- 10}$

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

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

The quadratic $8{x}^{2} + bx + c = 0,$

sum of roots $= \dfrac{- b}{8}$ and product of roots $= \dfrac{c}{8}$

$∴ c - b = 8 \times [$ (product of roots$) + ($sum of roots$) ]$

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