Find the value of,
\begin{aligned}
\frac{1}{216^{-\frac{2}{3}}}+\frac{1}{256^{-\frac{3}{4}}}+\frac{1}{32^{-\frac{1}{5}}}
\end{aligned}

Answer: Option B

Explanation:

\begin{align}&\left(\dfrac{a}{b}\right)^{x-2} = \left(\dfrac{b}{a}\right)^{x-7}\\\\
&\Rightarrow \left(\dfrac{a}{b}\right)^{x-2} = \left(\dfrac{a}{b}\right)^{-(x-7)}\\\\
&\Rightarrow x - 2 = -(x - 7)\\\\
&\Rightarrow x - 2 = -x + 7\\\\
&\Rightarrow x-2 = -x + 7\\\\
&\Rightarrow 2x = 9\\\\
&\Rightarrow x = \dfrac{9}{2} = 4.5
\end{align}

Answer: Option B

Explanation:

\begin{align}&\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)^2\\\\
&= x - 2 + \dfrac{1}{x}\\\\
&= x + \dfrac{1}{x} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{1}{\left(8 + 3\sqrt{7}\right)} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{\left(8 + 3\sqrt{7}\right)\left(8 - 3\sqrt{7}\right)} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{8^2 - \left(3\sqrt{7}\right)^2} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{64 - 63} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{1} - 2 \\\\
&= 8 + 3\sqrt{7} + 8 - 3\sqrt{7} - 2 \\\\
&= 14 \\\\
&\text{as }\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)^2 = 14\\\\
&\text{so ,}\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right) = \sqrt{14}\end{align}

Answer: Option C

Explanation:

\begin{aligned}
= \frac{(10^3)^7}{(10)^{18}}
\end{aligned}

\begin{aligned}
= \frac{(10)^{21}}{(10)^{18}} = 10^3 = 1000
\end{aligned}

Answer: Option D

Explanation:

\begin{aligned}3^{x-y} = 27 = 3^3 <=> x-y = 3 \text{... (i)}\\
3^{x+y} = 243 = 3^5 <=> x+y = 5 \text{... (ii)} \\

\text{ adding (i) and (ii)}
=> 2x = 8 \\
=> x = 4
\end{aligned}

Answer: Option C