Question : If $\left(y^2+\frac{1}{y^2}\right)=74$ and $y>1$, then find the value of $\left(y-\frac{1}{y}\right)$.

Option 1: $6 \sqrt{2}$

Option 2: $-2 \sqrt{19}$

Option 3: $2 \sqrt{19}$

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

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: $4 \sqrt{2}$

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 $\left(3 y+\frac{3}{y}\right)=8$, then find the value of $\left(y^2+\frac{1}{y^2}\right)$.

Option 1: $5\frac{1}{9}$

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

Option 3: $7\frac{1}{9}$

Option 4: $9\frac{1}{9}$

Question : If $a=\frac{1}{a-\sqrt{6}}$ and $(a>0)$, then the value of $\left(a+\frac{1}{a}\right)$ is:

Option 1: $\sqrt{6}$

Option 2: $\sqrt{10}$

Option 3: $\sqrt{15}$

Option 4: $\sqrt{7}$