Question

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

• A. $\pm \frac{23}{8}$
• B. $\pm \frac{73}{8}$
• C. $\pm \frac{63}{8}$
• D. $\pm \frac{13}{8}$

Answer 1

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

$2(x^2+1)=5x$

$=>\frac{x^2+1}{x}=\frac{5}{2}$

$=>x+\frac{1}{x}=\frac{5}{2}$

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,

$x^3-\frac{1}{x^3}=(x-\frac{1}{x})\left[\right(x^2+\frac{1}{x^2}+\left(x\right)\left(\frac{1}{x}\right)]$

$=(\pm \frac{3}{2})(\frac{17}{4}+1)$

$=(\pm \frac{3}{2})\left(\frac{21}{4}\right)$

$=\pm \frac{63}{8}$