1. \( x^{3}-3 x^{2}+2 x=0 \) 2. \( 3 x^{3}-2 x^{2}-8 x=0 \) 3. \( 6 x\left(x^{2}-36\right)=0 \)

Let's consider the first equation, \( x^{3}-3 x^{2}+2 x=0 \). To solve it, we can factor out an \( x \), giving us \( x(x^{2}-3x+2)=0 \). This further factors to \( x(x-1)(x-2)=0 \), leading to the solutions \( x=0, 1, 2 \). Now for the second equation, \( 3 x^{3}-2 x^{2}-8 x=0 \), we can factor out \( x \) as well, leaving us with \( x(3x^{2}-2x-8)=0 \). We now solve \( 3x^{2}-2x-8=0 \) using the quadratic formula \( x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \). Plugging in \( a=3, b=-2, c=-8 \), we find two additional solutions to the equation along with \( x=0 \). For the third equation, \( 6 x\left(x^{2}-36\right)=0 \), we can set each factor to zero: either \( 6x=0 \) (yielding \( x=0 \)) or \( x^{2}-36=0 \), which leads us to \( x=6 \) and \( x=-6 \). In total, you would find the roots for each of these equations to be \( x=0, 1, 2, 6, -6 \). Insights like these can help illuminate patterns in polynomial behavior and solutions!