d) \displaystyle{10}{x}^{{2}}{y}^{{3}} Explanation: \displaystyle={\sqrt[{{3}}]{{{20}{x}^{{2}}{y}^{{7}}}}}\cdot{\sqrt[{{3}}]{{{50}{x}^{{4}}{y}^{{2}}}}} \displaystyle={\sqrt[{{3}}]{{{\left({20}{x}^{{2}}{y}^{{7}}\right)}\cdot{\left({50}{x}^{{4}}{y}^{{2}}\right)}}}} ...

At the stage where you simplify \sqrt{(xy)^2} you should get |xy|, not xy. Therefore your final line should be |xy||x|, which is equal to |x|^2 |y| = x^2 |y|

For any complex number w = a + bj, |w| = \sqrt{a^2 + b^2} ww^* = |w|^2 = a^2 + b^2 hence if z=x + yj, then \begin{align*} z-j &= x +yj - j = x + (y-1)j\\[4pt] \implies\; ...

As mentioned in comments, it would be nice to show that the expression defines a real number for suitable x. If we truncate at various points, we get the expressions x^{1/3}, x^{1/3+1/9}, x^{1/3+1/9+1/27} ...

Dividing (or, in general, "combining") several equations can be used to derive new equations that must necessarily hold, but they are not equivalent to the original system. Consider for example: \begin{cases} x = 1 \\ y = 1 \end{cases} ...