Solve \[ x^{3}-9 x^{2}+324=36 x \]

To solve the equation \[ x^{3} - 9x^{2} + 324 = 36x, \] we first rearrange all terms to one side: \[ x^{3} - 9x^{2} - 36x + 324 = 0. \] Next, we can use the Rational Root Theorem or test for possible rational roots by substituting various integer values. Let's try \(x = 12\): \[ 12^{3} - 9(12^{2}) - 36(12) + 324. \] Calculating step-by-step: - \(12^{3} = 1728\) - \(9(12^{2}) = 9 \times 144 = 1296\) - \(36(12) = 432\) Now substituting these values into the polynomial: \[ 1728 - 1296 - 432 + 324 = 1728 - 1728 = 0. \] So, \(x = 12\) is indeed a root. Now, we can factor \(x - 12\) out of the cubic polynomial using synthetic division. Performing synthetic division of \(x^3 - 9x^2 - 36x + 324\) by \(x - 12\): \[ \begin{array}{r|rrrr} 12 & 1 & -9 & -36 & 324 \\ & & 12 & 36 & 0 \\ \hline & 1 & 3 & 0 & 324 \\ \end{array} \] The quotient is \(x^{2} + 3x + 27\). Therefore, we can express the cubic as: \[ (x - 12)(x^2 + 3x + 27) = 0. \] To find the remaining roots, we need to solve: \[ x^2 + 3x + 27 = 0. \] Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 3\), and \(c = 27\): Calculating the discriminant: \[ b^2 - 4ac = 3^2 - 4 \times 1 \times 27 = 9 - 108 = -99. \] Since the discriminant is negative, the quadratic has no real roots. Thus, we summarize the solution as follows: The real solution is \(x = 12\) and the other two roots are complex, found from \[ x = \frac{-3 \pm \sqrt{-99}}{2} = \frac{-3 \pm 3i\sqrt{11}}{2}. \] Final solutions: \[ \boxed{12}, \; \frac{-3 + 3i\sqrt{11}}{2}, \; \frac{-3 - 3i\sqrt{11}}{2}. \]