If x+frac{1}{x}=a, x^2+frac{1}{x^3}=b, then x | vedclass

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Question

(a)

Given, $x+\frac{1}{x}=a$ and $x^2+\frac{1}{x^3}=b$

Now, squaring both sides, we get

$\qquad\left(x+\frac{1}{x}\right)^2=a^2$

$\Rightar

Vedclass · Accepted Answer

$a^3+a^2-3 a-2-b$

Vedclass · Answer

(a)

Given, $x+\frac{1}{x}=a$ and $x^2+\frac{1}{x^3}=b$

Now, squaring both sides, we get

$\qquad\left(x+\frac{1}{x}ight)^2=a^2$

$\Rightarrow \quad x^2+\frac{1}{x^2}+2=a^2$

$	ext { On cubing both sides, we get }$

$\quad\left(x+\frac{1}{x}ight)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}ight)=a^3 \ldots 	ext { (ii) }$

$	ext { On adding Eqs. (i) and (ii), we get }$

$\left(x^2+\frac{1}{x^3}ight)+\left(x^3+\frac{1}{x^2}ight)+2+3\left(x+\frac{1}{x}ight)$

$\Rightarrow \quad b+\left(x^3+\frac{1}{x^2}ight)+2+3 a=a^3+a^2$

$x^3+\frac{1}{x^2}=a^3+a^2-3 a-b-2$

If $x+\frac{1}{x}=a, x^2+\frac{1}{x^3}=b$, then $x^3+\frac{1}{x^2}$ is

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