Question
If $2\left(x^2+1\right)=5 x$, find:$(i) x-\frac{1}{x};(ii) x^3-\frac{1}{x^3}$
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Answer
$\text { (i) } 2\left(x^2+1\right)=5 x$
$ \left(x^2+1\right)=\frac{5}{2} x$
Dividing by $\mathrm{x}$, we have
$\frac{x^2+1}{x}=\frac{5}{2}$
$\Rightarrow\left(x+\frac{1}{x}\right)=\frac{5}{2}\ldots . .(1)$
Now consider the expansion of $\left(x+\frac{1}{x}\right)^2$ :
$\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2$
$ \Rightarrow\left(\frac{5}{2}\right)^2=x^2+\frac{1}{x^2}+2$
$ \Rightarrow\left(\frac{5}{2}\right)^2-2=x^2+\frac{1}{x^2}$
$ \Rightarrow \frac{25}{4}-2=x^2+\frac{1}{x^2}$
$ \Rightarrow x^2+\frac{1}{x^2}=\frac{25-8}{4}[$From$(1)]$
$\Rightarrow x^2+\frac{1}{x^2}=\frac{17}{4}\ldots .(2)$
Now consider the expansion of $\left(x-\frac{1}{x}\right)^2$ :
$\left(x-\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}-2$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=\frac{17}{4}-2 [$from$(2)]$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=\frac{17-8}{4}$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=\frac{9}{4}$
$\Rightarrow\left(x-\frac{1}{x}\right)^2= \pm \frac{3}{2}\ldots .(3)$
$(ii)$ We know that,
$\left(x^3-\frac{1}{x^3}\right)=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$
$ \therefore\left(x^3-\frac{1}{x^3}\right)=\left( \pm \frac{3}{2}\right)^3+3\left( \pm \frac{3}{2}\right)[$from$(3)]$
$ = \pm \frac{27}{8}+\frac{9}{2}$
$ \Rightarrow\left(x^3-\frac{1}{x^3}\right)= \pm \frac{27+36}{8}$
$ \Rightarrow\left(x^3-\frac{1}{x^3}\right)= \pm \frac{63}{8}$
$ \left(x^2+1\right)=\frac{5}{2} x$
Dividing by $\mathrm{x}$, we have
$\frac{x^2+1}{x}=\frac{5}{2}$
$\Rightarrow\left(x+\frac{1}{x}\right)=\frac{5}{2}\ldots . .(1)$
Now consider the expansion of $\left(x+\frac{1}{x}\right)^2$ :
$\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2$
$ \Rightarrow\left(\frac{5}{2}\right)^2=x^2+\frac{1}{x^2}+2$
$ \Rightarrow\left(\frac{5}{2}\right)^2-2=x^2+\frac{1}{x^2}$
$ \Rightarrow \frac{25}{4}-2=x^2+\frac{1}{x^2}$
$ \Rightarrow x^2+\frac{1}{x^2}=\frac{25-8}{4}[$From$(1)]$
$\Rightarrow x^2+\frac{1}{x^2}=\frac{17}{4}\ldots .(2)$
Now consider the expansion of $\left(x-\frac{1}{x}\right)^2$ :
$\left(x-\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}-2$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=\frac{17}{4}-2 [$from$(2)]$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=\frac{17-8}{4}$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=\frac{9}{4}$
$\Rightarrow\left(x-\frac{1}{x}\right)^2= \pm \frac{3}{2}\ldots .(3)$
$(ii)$ We know that,
$\left(x^3-\frac{1}{x^3}\right)=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$
$ \therefore\left(x^3-\frac{1}{x^3}\right)=\left( \pm \frac{3}{2}\right)^3+3\left( \pm \frac{3}{2}\right)[$from$(3)]$
$ = \pm \frac{27}{8}+\frac{9}{2}$
$ \Rightarrow\left(x^3-\frac{1}{x^3}\right)= \pm \frac{27+36}{8}$
$ \Rightarrow\left(x^3-\frac{1}{x^3}\right)= \pm \frac{63}{8}$
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