How do people come up with solutions like this? I was going through some problems in high school textbook and stumbled on this problem.
I could be trying to solve it whole day and I wouldn't solve it. The solution seems to be too complicated for high school student (I marked the part red I can't understand). Can you explain how do you go about solving this problem and what is going through your mind when doing so?

 A: One problem comprehending such arguments is  inessential information is obfuscating the algebra, e.g. the fractional exponents. They have a common denominator of $3$ so writing $\,x = a^{1/3},\ y = b^{1/3}$ simplifies it, making the steps much clearer, namely
$$\dfrac{-2x^3y^4 + 2 x^5y^2}{x^6y^4 - y^6 x^4} = \dfrac{2x^3y^2 (\color{#c00}{-y^2+x^2})}{x^4 y^4\, (\ \color{#c00}{x^2\,-\,y^2})} = \dfrac{2}{xy^2} = \dfrac{2}{a^{1/3}b^{2/3}} $$
A: Through practice, you become more familiar with various techniques which you can use to make simplification easier and faster.
The previous answer provides a clever substitution which simplifies the problem greatly, but you can solve the question rather easily even without it. In the example, you have
$$\frac{-2ab^{\frac{4}{3}}+2a^{\frac{5}{3}}b^{\frac{2}{3}}}{a^2b^{\frac{4}{3}}-b^2a^{\frac{4}{3}}}$$
In order to understand the steps given for the solution, you can break the problem down into two parts. Focusing on the numerator first, you can rewrite it as
$$-2abb^{\frac{1}{3}}+2aa^{\frac{2}{3}}b^{\frac{2}{3}}$$
because we have $a^{b+c} = a^ba^c$, so, for instance $b^{\frac{4}{3}} = b^{1+\frac{1}{3}} = bb^{\frac{1}{3}}$. Using this technique again, you can split $b$ into $b^{\frac{1}{3}}b^{\frac{2}{3}}$, so you get
$$-\color{blue}{2ab^{\frac{2}{3}}}b^{\frac{1}{3}}b^{\frac{1}{3}}+\color{blue}{2a}a^{\frac{2}{3}}\color{blue}{b^{\frac{2}{3}}}$$
Notice how both sides share common factors. You have $ab+ac = a(b+c)$, so the expression can once more be rewritten as
$$2ab^{\frac{2}{3}}\left(-b^{\frac{2}{3}}+a^{\frac{2}{3}}\right) \tag{1}$$
In fact, in the picture you’ve given, the simplification wasn’t immediately made. Rather, all the steps were shown, as in $ab+ac = a\left(\frac{ab}{a}+\frac{ac}{a}\right)$ rather than immediately $ab+ac = a(b+c)$.
Repeat the same process for the denominator (rewriting and factoring):
$$a^2b^{\frac{4}{3}}-b^2a^{\frac{4}{3}} = a^{\frac{2}{3}}\color{blue}{a^{\frac{4}{3}}b^{\frac{4}{3}}}-b^{\frac{2}{3}}\color{blue}{b^{\frac{4}{3}}a^{\frac{4}{3}}} = a^{\frac{4}{3}}b^{\frac{4}{3}}\left(a^{\frac{2}{3}}-b^{\frac{2}{3}}\right) \tag{2}$$
Putting it all together, you reach
$$\frac{2ab^{\frac{2}{3}}\color{blue}{\left(-b^{\frac{2}{3}}+a^{\frac{2}{3}}\right)}}{a^{\frac{4}{3}}b^{\frac{4}{3}}\color{blue}{\left(a^{\frac{2}{3}}-b^{\frac{2}{3}}\right)}} = \frac{2ab^{\frac{2}{3}}}{a^{\frac{4}{3}}b^{\frac{4}{3}}} = \frac{2ab^{\frac{2}{3}}}{aa^{\frac{1}{2}}b^{\frac{2}{3}}b^{\frac{2}{3}}} = \frac{2}{a^{\frac{1}{3}}b^{\frac{2}{3}}}$$
As you can see, there really was nothing apart from manipulating the exponents and factoring. At first, this might seem long, tricky, and time-consuming. You need some practice in order to see these ideas and patterns quickly, and you’ll eventually solve these problems very quickly.
