If a = 2^x and b = 2^{x + 1}; show that :

$\dfrac{8a^3}{b^2} = 2^{x+1}$.

Given: a = 2^x and b = 2^{x + 1}

Now, $\dfrac{8a^3}{b^2} = 2^{x+1}$

Taking L.H.S.:

$= \dfrac{8a^3}{b^2}\\[1em] = \dfrac{8 \times (2^x)^3}{(2^{x + 1})^2}\\[1em] = \dfrac{2^3 \times 2^{3x}}{(2^{2x + 2})}\\[1em] = 2^3 \times 2^{3x - 2x - 2}\\[1em] = 2^{3 + x - 2}\\[1em] = 2^{x + 1}$

We have, R.H.S. = $2^{x+1}$

Thus, L.H.S. = R.H.S.

Hence, $\dfrac{8a^3}{b^2} = 2^{x+1}$.