नुहोस् (If \( 2 t=5-\sqrt{21} \) then, prove that) : \( t^{3}+\frac{1}{t^{3}}= \) होस् (If \( t=\sqrt{3+\sqrt{5}} \) then, prove that) : \( t^{2}+\frac{4}{t^{2}}=6 \)

Let's solve the two problems one by one. ### Problem 1: Prove that if \( 2t = 5 - \sqrt{21} \), then \( t^3 + \frac{1}{t^3} = ? \) 1. First, we can express \( t \) in terms of \( \sqrt{21} \): \[ t = \frac{5 - \sqrt{21}}{2} \] 2. Next, we need to find \( t^3 + \frac{1}{t^3} \). We can use the identity: \[ t^3 + \frac{1}{t^3} = \left(t + \frac{1}{t}\right)^3 - 3\left(t + \frac{1}{t}\right) \] 3. To use this identity, we first need to find \( t + \frac{1}{t} \): \[ \frac{1}{t} = \frac{2}{5 - \sqrt{21}} \] To rationalize the denominator: \[ \frac{1}{t} = \frac{2(5 + \sqrt{21})}{(5 - \sqrt{21})(5 + \sqrt{21})} = \frac{2(5 + \sqrt{21})}{25 - 21} = \frac{2(5 + \sqrt{21})}{4} = \frac{5 + \sqrt{21}}{2} \] 4. Now, we can find \( t + \frac{1}{t} \): \[ t + \frac{1}{t} = \frac{5 - \sqrt{21}}{2} + \frac{5 + \sqrt{21}}{2} = \frac{(5 - \sqrt{21}) + (5 + \sqrt{21})}{2} = \frac{10}{2} = 5 \] 5. Now we can substitute \( t + \frac{1}{t} \) into the identity: \[ t^3 + \frac{1}{t^3} = 5^3 - 3 \cdot 5 = 125 - 15 = 110 \] Thus, we have: \[ t^3 + \frac{1}{t^3} = 110 \] ### Problem 2: Prove that if \( t = \sqrt{3 + \sqrt{5}} \), then \( t^2 + \frac{4}{t^2} = 6 \) 1. First, we calculate \( t^2 \): \[ t^2 = 3 + \sqrt{5} \] 2. Next, we need to find \( \frac{4}{t^2} \): \[ \frac{4}{t^2} = \frac{4}{3 + \sqrt{5}} \] To rationalize the denominator: \[ \frac{4}{t^2} = \frac{4(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} = \frac{4(3 - \sqrt{5})}{9 - 5} = \frac{4(3 - \sqrt{5})}{4} = 3 - \sqrt{5} \] 3. Now we can find \( t^2 + \frac{4}{t^2} \): \[ t^2 + \frac{4}{t^2} = (3 + \sqrt{5}) + (3 - \sqrt{5}) = 3 + \sqrt{5} + 3 - \sqrt{5} = 6 \] Thus, we have: \[ t^2 + \frac{4}{t^2} = 6 \] ### Summary of Results 1. For \( 2t = 5 - \sqrt{21} \), we proved that \( t^3 + \frac{1}{t^3} = 110 \). 2. For \( t = \sqrt{3 + \sqrt{5}} \), we proved that \( t^2 + \frac{4}{t^2} = 6 \).