Practicing Success
Practicing Success
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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? |
2 256 0 128 |
2 |
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 |
