b) If \( 6 x-\frac{1}{6 x}=6 \), then find the value of \( 216 x^{3}-\frac{1}{216 x^{3}} \)

To solve the equation \( 6x - \frac{1}{6x} = 6 \), let's first multiply the entire equation by \( 6x \) to eliminate the fraction: \[ (6x)(6x) - (1) = 6(6x) \] This simplifies to: \[ 36x^2 - 1 = 36x \] Rearranging gives us a standard quadratic equation: \[ 36x^2 - 36x - 1 = 0 \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 36 \), \( b = -36 \), and \( c = -1 \): \[ b^2 - 4ac = (-36)^2 - 4(36)(-1) = 1296 + 144 = 1440 \] Thus, the solutions for \( x \) are: \[ x = \frac{36 \pm \sqrt{1440}}{72} = \frac{36 \pm 12\sqrt{10}}{72} = \frac{1 \pm \frac{\sqrt{10}}{3}}{2} \] Next, we want to find \( 216x^3 - \frac{1}{216x^3} \). To do this, we will first determine \( x + \frac{1}{6x} \): Let \( y = 6x \), then we have: \[ y - \frac{1}{y} = 6 \] Rearranging this gives: \[ y^2 - 6y - 1 = 0 \] We can use the quadratic formula again for \( y \): \[ y = \frac{6 \pm \sqrt{36 + 4}}{2} = \frac{6 \pm \sqrt{40}}{2} = 3 \pm \sqrt{10} \] Thus, we have: \[ 6x = 3 \pm \sqrt{10} \implies x = \frac{3 \pm \sqrt{10}}{6} \] Now, we'll use: \[ z = x + \frac{1}{6x} = \frac{3 \pm \sqrt{10}}{6} + \frac{1}{3 \pm \sqrt{10}} \] Calculating: \[ \frac{1}{3 + \sqrt{10}} = \frac{3 - \sqrt{10}}{(3+\sqrt{10})(3-\sqrt{10})} = \frac{3 - \sqrt{10}}{-1} = \sqrt{10} - 3 \] So: \[ y + \frac{1}{y} = \left(3 + \sqrt{10}\right) + \left(\sqrt{10} - 3\right) = 2\sqrt{10} \] Now we find \( x^3 + \frac{1}{6^3x^3} \): Using the identity: \[ x^3 + \frac{1}{6^3x^3} = (x + \frac{1}{6x})^3 - 3(x + \frac{1}{6x}) \] Setting \( a = y + \frac{1}{y} = 2\sqrt{10} \), we find: \[ (x + \frac{1}{6x})^3 = 8\sqrt{10}^3 = 80\sqrt{10} \] Thus: \[ 216x^3 - \frac{1}{216x^3} = \left(2\sqrt{10}\right)^3 - 3(2\sqrt{10}) = 80\sqrt{10} - 6\sqrt{10} = 74\sqrt{10} \] Finally, we have: \[ \boxed{74\sqrt{10}} \]