If $x+\frac{1}{x}=a, x^2+\frac{1}{x^3}=b$, then $x^3+\frac{1}{x^2}$ is
If $x+\frac{1}{x}=a, x^2+\frac{1}{x^3}=b$, then $x^3+\frac{1}{x^2}$ is
- [KVPY 2011]
- A
$a^3+a^2-3 a-2-b$
- B
$a^3-a^2-3 a+4-b$
- C
$a^3-a^2+3 a-6-b$
- D
$a^3+a^2+3 a-16-b$
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$II$. There are infinitely many values of $n$ for which both roots are real.
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