Question

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

• A. $\frac{83}{206}$
• B. $\frac{91}{216}$
• C. $\frac{71}{206}$
• D. $\frac{97}{236}$

Answer 1

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

$x^2+\frac{1}{9x^2}=\frac{25}{36}$

We know,

$(x+\frac{1}{3x}{)}^2=x^2+2(x\left)\right(\frac{1}{3x})+(\frac{1}{9x^2})$

$=\frac{25}{36}+\frac{2}{3}$

$=\frac{25+24}{36}$

$=\frac{49}{36}$

$=>(x+\frac{1}{3x})=\sqrt{\frac{49}{36}}$

$=>x+\frac{1}{3x}=\pm \frac{7}{6}$

Now,

$x^3+\frac{1}{27x^3}=(x+\frac{1}{3x})\left[\right(x{)}^2+(\frac{1}{9x}{)}^2-(x\left)\right(\frac{1}{3x}\left)\right]$

$=(\pm \frac{7}{6})(\frac{25}{36}-\frac{1}{3})$

$=(\pm \frac{7}{6})\left(\frac{25-12}{36}\right)$

$=(\pm \frac{7}{6})\left(\frac{13}{36}\right)$

$=(\pm \frac{91}{216})$