If \(a-\displaystyle\frac{1}{a-5}=16\), then find the value of \((a-5)^2+\displaystyle\frac{1}{(a-5)^2}\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 123
Subtracting 5 from both sides makes the left side \(t-\frac{1}{t}\) with \(t=a-5\). Squaring uses \(\left(t-\frac{1}{t}\right)^2=t^2+\frac{1}{t^2}-2\).
\((a-5)-\displaystyle\frac{1}{a-5}=16-5=11\)
\((a-5)^2+\displaystyle\frac{1}{(a-5)^2}=11^2+2\)
\(=121+2=123\)