Question : The simplified value of $\frac{3\sqrt 2 }{\sqrt3 + \sqrt6} - \frac{4 \sqrt 3 }{\sqrt{6}+ \sqrt {2}} + \frac{\sqrt 6}{\sqrt{3}+ \sqrt 2}$ is:

Option 1: $\sqrt 2$

Option 2: $\frac{1}{\sqrt2}$

Option 3: $\sqrt 3- \sqrt 2$

Option 4: $0$

Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:

Option 1: $\frac{2 \sqrt{6}-6}{5}$

Option 2: $\frac{4 \sqrt{6}-6}{5}$

Option 3: $\frac{4 \sqrt{6}-6}{3}$

Option 4: $\frac{4 \sqrt{3}-6}{5}$

Question : The value of $\frac{1}{1+\sqrt{2}+\sqrt{3}}+\frac{1}{1-\sqrt{2}+\sqrt{3}}$ is:

Option 1: $\sqrt{2}$

Option 2: $\sqrt{3}$

Option 3: $1$

Option 4: $4(\sqrt{3}+\sqrt{2})$

Question : If $x=\sqrt{3}-\frac{1}{\sqrt{3}}, y=\sqrt{3}+\frac{1}{\sqrt{3}}$, then the value of $\frac{x^2}{y}+\frac{y^2}{x}$ is:

Option 1: $\sqrt{3}$

Option 2: $3\sqrt{3}$

Option 3: $16\sqrt{3}$

Option 4: $2\sqrt{3}$

Question : The value of $x$ in the expression $\tan^{2}\frac{\pi }{4}-\cos^{2}\frac{\pi }{3}=x\sin\frac{\pi }{4}\cos\frac{\pi }{4}\tan\frac{\pi }{3}$ is:

Option 1: $\frac{2}{\sqrt{3}}$

Option 2: $\frac{3\sqrt{3}}{4}$

Option 3: $\frac{1}{\sqrt{3}}$

Option 4: $\frac{\sqrt{3}}{2}$