Question

If $2(x^2+1)=5x$, find $x^3-\frac{1}{x^3}$.

• A. $\pm \frac{17}{3}$
• B. $\pm \frac{63}{8}$
• C. $\pm \frac{13}{8}$
• D. $\pm \frac{3}{8}$

Answer 1

The correct option is B $\pm \frac{63}{8}$

$2(x^2+1)=5x$
$=>\frac{x^2+1}{x}=\frac{5}{2}$
$=>x+\frac{1}{x}=\frac{5}{2}$
Squaring both sides,
$=>x^2+\frac{1}{x^2}+2\left(x\right)+\left(\frac{1}{x}\right)=\frac{25}{4}$
$=>x^2+\frac{1}{x^2}=\frac{25}{4}-2$
$=>x^2+\frac{1}{x^2}=\frac{17}{4}$
Now,
$(x-\frac{1}{x^2}{)}^2=x^2+\frac{1}{x^2}-2(x)+(\frac{1}{x})$
$=\frac{17}{4}-2$
$=\frac{9}{4}$
$=>x-\frac{1}{x}=\pm \sqrt{\frac{9}{4}}$
$=\pm \frac{3}{2}$
Now,
$x^3-\frac{1}{x^3}=(x-\frac{1}{x})\left[\right(x^2+\frac{1}{x^2}+\left(x\right)\left(\frac{1}{x}\right)]$
$=(\pm \frac{3}{2})(\frac{17}{4}+1)$
$=(\pm \frac{3}{2})\left(\frac{21}{4}\right)$
$=\pm \frac{63}{8}$