If \(a+\displaystyle\frac{1}{a+2}=4\), then find the value of \((a+2)^3+\displaystyle\frac{1}{(a+2)^3}\).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 198
Adding 2 to both sides makes the left side \(t+\frac{1}{t}\) with \(t=a+2\). The cube then follows from \(t^3+\frac{1}{t^3}=\left(t+\frac{1}{t}\right)^3-3\left(t+\frac{1}{t}\right)\).
\((a+2)+\displaystyle\frac{1}{a+2}=4+2=6\)
\((a+2)^3+\displaystyle\frac{1}{(a+2)^3}=6^3-3\times6\)
\(=216-18=198\)