If \(a+\displaystyle\frac{1}{a}+2=0\), then find the value of \(a^{36}-\displaystyle\frac{1}{a^{35}}\).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
Multiplying the condition by \(a\) turns it into a perfect square, which fixes \(a\) exactly. Then an even power of \(-1\) is \(1\) and an odd power is \(-1\).
\(a^2+1+2a=0\Rightarrow(a+1)^2=0\), so \(a=-1\)
\(a^{36}=(-1)^{36}=1\) (even power)
\(a^{35}=(-1)^{35}=-1\) (odd power), so \(\displaystyle\frac{1}{a^{35}}=-1\)
\(a^{36}-\displaystyle\frac{1}{a^{35}}=1-(-1)=2\)