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 : If $\left(x+\frac{1}{x}\right)=5 \sqrt{2}$, and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right) ?$

Option 1: $22970 \sqrt{23}$

Option 2: $23030 \sqrt{23}$

Option 3: $23060 \sqrt{23}$

Option 4: $22960 \sqrt{23}$

Question : If $x=\frac{1}{x-5}(x>0)$, then the value of $x+\frac{1}{x}$ is:

Option 1: $\sqrt{41}$

Option 2: $\sqrt{29}$

Option 3: $\sqrt{23}$

Option 4: $\sqrt{43}$

Question : If $a+b :\sqrt{ab} = 4:1 $ where $ a > b > 0$, then $ a:b$ is:

Option 1: $\left (2+\sqrt{3} \right):\left (2-\sqrt{3} \right)$

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

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

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

Question : If $x^3=270+y^3$ and $x=(6+y)$, then what is the value of $(x+y)? $(given that $x>0$ and $y>0$)

Option 1: $2 \sqrt{3}$

Option 2: $\sqrt{3}$

Option 3: $4 \sqrt{3}$

Option 4: $4 \sqrt{2}$