Question

The value of $\frac{\left\{{\left(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}\right)}^2\left(\sqrt{3}-\sqrt{2}\right)\right\}}{{\left(\sqrt{\sqrt{3}+1}\right)}^2-{\left(\sqrt{\sqrt{3}-1}\right)}^2}$ is

• A. $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}}$
• B. $1$
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is B

$1$

Explanation for the correct answer:

Compute the required value:

$\frac{\left\{{\left(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}\right)}^2\left(\sqrt{3}-\sqrt{2}\right)\right\}}{{\left(\sqrt{\sqrt{3}+1}\right)}^2-{\left(\sqrt{\sqrt{3}-1}\right)}^2}=\frac{\left(\sqrt{3}+1+\sqrt{3}-1+2\sqrt{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}+1-\sqrt{3}-1}$ $...(\because {\left(a+b\right)}^2=a^2+2ab+b^2)$

$=\frac{\left(2\sqrt{3}+2\sqrt{3-1}\right)\left(\sqrt{3}-\sqrt{2}\right)}{2}$ $...\left(\because \left(a^2-b^2\right)=(a+b)(a-b)\right)$

$=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)$

$=3-2$

$=1$

Hence, the value of $\frac{\left\{{\left(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}\right)}^2\left(\sqrt{3}-\sqrt{2}\right)\right\}}{{\left(\sqrt{\sqrt{3}+1}\right)}^2-{\left(\sqrt{\sqrt{3}-1}\right)}^2}$ is $1.$

Hence. option $\left(B\right)$ is the correct answer.