If \(a^2+b^2+\displaystyle\frac{1}{a^2}+\frac{1}{b^2}=4\), where \(a\neq0\) and \(b\neq0\), then the value of \(a^{256}+b^{256}\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
For any non-zero \(t\), \(t^2+\frac{1}{t^2}-2=\left(t-\frac{1}{t}\right)^2\ge0\), so \(t^2+\frac{1}{t^2}\) is never less than 2. Two such sums can add up to 4 only when each of them is exactly 2.
\(\left(a^2+\displaystyle\frac{1}{a^2}-2\right)+\left(b^2+\frac{1}{b^2}-2\right)=0\)
\(\left(a-\displaystyle\frac{1}{a}\right)^2+\left(b-\frac{1}{b}\right)^2=0\)
So \(a=\displaystyle\frac{1}{a}\) and \(b=\displaystyle\frac{1}{b}\), that is \(a^2=1\) and \(b^2=1\)
\(a^{256}=\left(a^2\right)^{128}=1\) and \(b^{256}=1\)
\(a^{256}+b^{256}=1+1=2\)