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

Given,

$\Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{25}{36} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 - \dfrac{2}{3} = \dfrac{25}{36} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 = \dfrac{25}{36} + \dfrac{2}{3} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 = \dfrac{25 + 24}{36} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 = \dfrac{49}{36}\\[1em] \Rightarrow x + \dfrac{1}{3x} = \sqrt{\dfrac{49}{36}} \\[1em] \Rightarrow x + \dfrac{1}{3x} = \pm \dfrac{7}{6}.$

Since, x is > 0,

∴ $x + \dfrac{1}{3x} = \dfrac{7}{6}$

By formula,

$\Rightarrow \Big(x^3 + \dfrac{1}{27x^3}\Big) = \Big(x + \dfrac{1}{3x}\Big)^3 - \Big(x + \dfrac{1}{3x}\Big)$ …….(1)

Substituting $x + \dfrac{1}{3x} = \dfrac{7}{6}$ in equation (1), we get :

$\Rightarrow \Big(x^3 + \dfrac{1}{27x^3}\Big) = \Big(\dfrac{7}{6}\Big)^3 - \Big(\dfrac{7}{6}\Big) \\[1em] = \dfrac{343}{216} - \dfrac{7}{6} \\[1em] = \dfrac{343 - 252}{216} \\[1em] = \dfrac{91}{216}.$

Hence, $x^3 + \dfrac{1}{27x^3} = \dfrac{91}{216}$.