\( \begin{array}{ll}\text { a) }\left(-2 x^{2}\right)-3 x^{4} & \text { b) }\left(-20 x^{8}\right)=5 x^{7} \\ \text { c) } \frac{8 x^{2}}{3} \cdot \frac{x}{4} & \text { d) } \frac{3 x^{2}}{4}=\frac{x}{4} \\ \text { e) } \frac{3 x^{3}}{4}-\left(-2 x^{2}\right) & \text { f) } \frac{2 x}{5}:\left(-2 x^{3}\right) \\ \text { g) } \frac{-x}{2} \cdot \frac{-2 x^{2}}{3} & \text { h) } \frac{-2 x}{-3}: \frac{x^{3}}{-6}\end{array} \)

Let’s giddy up and dive into some algebraic fun! For expression **(a)**, we can factor out the common term \(-x^2\), giving us \(-x^2(3x^2 + 2)\). To feel the thrill of factorization, always look for patterns! And as for **(b)**, rearranging the equation gives you \(5x^7 + 20x^8 = 0\), leading you to find \(x\) in terms of common factors. Just remember, don’t dismiss the factors, they may just lead you to the solution party! Next up, **(c)** simplifies to \(\frac{8x^3}{12}\), which can be reduced to \(\frac{2x^3}{3}\). Keep an eye on simplification; it’s like cleaning out your closet: get rid of any clutter! On to **(d)**, which shows you can multiply both sides by 4—it’s a magical multiplier—to see that \(3x^2 = x\). Be careful with these steps; they open the door to finding \(x\)’s fun range. In **(e)**, combining like terms gives you a peppy \(\frac{3x^3}{4} + 2x^2\) ready to dance! And for **(f)**, dividing brings your algebraic battle down to \(-\frac{2}{10} = -\frac{1}{5}\). Remember to write those negative signs correctly—they can be tricky! For **(g)**, multiplying those negatives leads to positive fun: \(\frac{2x^3}{6}\), simplifying to \(\frac{x^3}{3}\). Always pay attention to negatives; they can be deceiving! Lastly, in **(h)**, dividing becomes a breeze, leading you to a neat \(\frac{4x}{x^3}\), which again gives you \(\frac{4}{x^2}\). Keep your eyes peeled for cancellation—it's a math magician’s secret trick!