If m and n are whole numbers such that
$$\begin{aligned} m^n=121 \end{aligned}$$
, the value of $$\begin{aligned} (m-1)^{n + 1} \end{aligned}$$ is

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:

$$\begin{aligned} 6^{m-2}\\ = \dfrac{6^m}{6^2}\\ = \dfrac{46656}{6^2}\\ = \dfrac{46656}{36} = 1296 \end{aligned}$$

Answer: Option B

Explanation:

Clue:

$$\begin{aligned} (\sqrt{x} - \frac{1}{\sqrt{x}})^2 = x + \frac{1}{x} - 2 \ \end{aligned}$$
Now put the value of x to calculate the answer :)

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 C