Question

If $x>0$ and $x^2+\frac{1}{9x^2}=\frac{25}{36}$, find : $x^3+\frac{1}{27x^3}$

• A. $\frac{220}{176}$
• B. $\frac{106}{517}$
• C. $\frac{91}{216}$
• D. $\frac{32}{71}$

Answer 1

The correct option is C $\frac{91}{216}$

$x^2+\frac{1}{9x^2}=\frac{25}{36}$
We know,
$(x+\frac{1}{3x}{)}^2=x^2+2(x\left)\right(\frac{1}{3x})+(\frac{1}{9x^2})$
$=\frac{25}{36}+\frac{2}{3}$
$=\frac{25+24}{36}$
$=\frac{49}{36}$
$=>(x+\frac{1}{3x})=\sqrt{\frac{49}{36}}$
$=>x+\frac{1}{3x}=\pm \frac{7}{6}$
since, $x$ is positive,
$=>x+\frac{1}{3x}=\frac{7}{6}$
Now,
$x^3+\frac{1}{27x^3}=(x+\frac{1}{3x})\left[\right(x{)}^2+(\frac{1}{9x}{)}^2-(x\left)\right(\frac{1}{3x}\left)\right]$
$=\left(\frac{7}{6}\right)(\frac{25}{36}-\frac{1}{3})$
$=\left(\frac{7}{6}\right)\left(\frac{25-12}{36}\right)$
$=\left(\frac{7}{6}\right)\left(\frac{13}{36}\right)$
$=\left(\frac{91}{216}\right)$