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4-2.Quadratic Equations and Inequations
hard
If $x+\frac{1}{x}=a, x^2+\frac{1}{x^3}=b$, then $x^3+\frac{1}{x^2}$ is
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
$a^3+a^2+3 a-16-b$
(KVPY-2011)
Solution
(a)
Given, $x+\frac{1}{x}=a$ and $x^2+\frac{1}{x^3}=b$
Now, squaring both sides, we get
$\qquad\left(x+\frac{1}{x}\right)^2=a^2$
$\Rightarrow \quad x^2+\frac{1}{x^2}+2=a^2$
$\text { On cubing both sides, we get }$
$\quad\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=a^3 \ldots \text { (ii) }$
$\text { On adding Eqs. (i) and (ii), we get }$
$\left(x^2+\frac{1}{x^3}\right)+\left(x^3+\frac{1}{x^2}\right)+2+3\left(x+\frac{1}{x}\right)$
$\Rightarrow \quad b+\left(x^3+\frac{1}{x^2}\right)+2+3 a=a^3+a^2$
$x^3+\frac{1}{x^2}=a^3+a^2-3 a-b-2$
Standard 11
Mathematics
