I have to compute the existence conditions of the following expression: $$\{\frac{3a^2}{x+2a}+[x^2:(1-\frac{a}{(x+a)})+a^2:(1-\frac{x}{(x+a)})]:\frac{(x^2+2ax+a^2)}{(x-2a)}\}:(\frac{(x^2-a^2)}{x^2}:\frac{(x^2-4a^2)}{x^2})=$$ $$=\{\frac{3a^2}{\color{red}{x+2a}}+[x^2:(\frac{\color{pink}x}{\color{green}{x+a}})+a^2:(\frac{\color{brown}a}{\color{green}{x+a}})]:\frac{(\color{violet}{x^2+2ax+a^2})}{(x-2a)}\}:(\frac{(\color{lightblue}{x^2-a^2})}{\bbox[yellow]{x^2}}:\frac{(\color{orange}{x^2-4a^2})}{\bbox[yellow]{x^2}})$$ Since all the fractions that appear have sense iff the denominator is zero and since I have to do some divisions ":", so I have thought: $$\begin{cases} \color{red}{x+2a}\neq 0\\ \color{green}{x+a}\neq 0\\ \color{pink}x\neq 0\\ \color{brown}a\neq 0\\ \color{violet}{x^2+2ax+a^2}\neq 0 \\ x-2a\neq 0\\ \bbox[yellow]{x^2}\neq 0\\ \color{lightblue}{x^2-a^2}\neq 0\\\color{orange}{x^2-4a^2}\neq 0 \\ (\frac{(\color{lightblue}{x^2-a^2})}{\bbox[yellow]{x^2}}:\frac{(\color{orange}{x^2-4a^2})}{\bbox[yellow]{x^2}})\neq 0\iff (\frac{(\color{lightblue}{x^2-a^2})}{\bbox[yellow]{x^2}}\cdot \frac{(\bbox[yellow]{x^2})}{\color{orange}{x^2-4a^2}})\iff \color{lightblue}{x^2-a^2}\neq 0 \end{cases}$$ And so$$ \begin{cases}x\neq -2a\\ x\neq -a\\ x\neq 0\\ a\neq 0\\ x\neq 2a\\ x\neq a\\\end{cases}$$
Do you think my work is well done? Please can you help me? I don't know where I can check my work...