If \(a-\displaystyle\frac{1}{a-3}=8\), then find the value of \((a-3)^2+\displaystyle\frac{1}{(a-3)^2}\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 27
Subtracting 3 from both sides makes the left side of the form \(t-\frac{1}{t}\) with \(t=a-3\); then \(t^2+\frac{1}{t^2}=\left(t-\frac{1}{t}\right)^2+2\).
\((a-3)-\displaystyle\frac{1}{a-3}=8-3=5\)
\((a-3)^2+\displaystyle\frac{1}{(a-3)^2}=5^2+2=25+2=27\)