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 C

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 A

Explanation:

$$\begin{aligned} = \frac{1}{\left( 1 + \frac{a^n}{a^m} \right)} + \frac{1}{\left( 1 + \frac{a^m}{a^n} \right)} \\ = \frac{a^m}{(a^m+a^n)} + \frac{a^n}{(a^m+a^n)} \\ = \frac{(a^m+a^n)}{(a^m+a^n)} = 1 \end{aligned}$$

Answer: Option D

Explanation:

$$\begin{aligned} = \frac{(10)^{150}}{(10)^{146}} = 10^4 = 10000 \end{aligned}$$

Answer: Option D

Explanation:

$$\begin{aligned}3^{x-y} = 27 = 3^3 <=> x-y = 3 \text{... (i)}\\ 3^{x+y} = 243 = 3^5 <=> x+y = 5 \text{... (ii)} \\ \text{ adding (i) and (ii)} => 2x = 8 \\ => x = 4 \end{aligned}$$