If a2 + b2 + \(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) = 4, a ≠ 0, b ≠ 0, then the value of \( {a }^{ 256} \) + \( {b }^{ 256} \) is?
Answer & explanation
Correct answer: option 1
Put a, b = 1 (let)
12 + 12 + \(\frac{1}{1^2}\) + \(\frac{1}{1^2}\) = 4 (satisfied)
Therefore, \( {a }^{ 256} \) + \( {b }^{ 256} \) = 1 + 1 = 2