If \(a-\displaystyle\frac{1}{a-9}=26\), then find the value of \((a-9)^2+\displaystyle\frac{1}{(a-9)^2}\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 291
Subtracting 9 from both sides makes the left side \(t-\frac{1}{t}\) with \(t=a-9\). Squaring uses \(\left(t-\frac{1}{t}\right)^2=t^2+\frac{1}{t^2}-2\).
\((a-9)-\displaystyle\frac{1}{a-9}=26-9=17\)
\((a-9)^2+\displaystyle\frac{1}{(a-9)^2}=17^2+2\)
\(=289+2=291\)