Question : If $A+\frac{1}{A}=-1$, then find the value of $\frac{{A}^6+{A}^3-1}{{~A}^9+{A}^3-1}$.
Option 1: 0
Option 2: –1
Option 3: 2
Option 4: 1
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Correct Answer: 1
Solution :
$A+\frac{1}{A}=-1$ ⇒ $A^2+A+1=0$
Now,
$A^3-1^3 = (A-1) (A^2+A+1)$
⇒$A^3-1 = (A-1)\times 0$
⇒$A^3-1 =0$
⇒$A^3=1$
So, $\frac{{A}^6+{A}^3-1}{{~A}^9+{A}^3-1}$
$=\frac{{(A^3)}^2+{A}^3-1}{{(A^3)}^3+{A}^3-1}$
$=\frac{1^2+1-1}{1^3+1-1}$
$= 1$
Hence, the correct answer is 1.
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