<s>[INST] Solving inequalities, simplifying radicals, and factoring. (Pre calculus) (Q.1) Solve for $x$ in $x^3 - 5x > 4x^2$
its a question in pre calculus for dummies workbook, chapter 2. The answer says:
then factor the quadratic: $x(x-5)(x+1)>0$. Set your factors equal to $0$ so you can find your key points.When you have them, put these points on a number line and plug in test numbers form each possible section to determine whether the factor would be positive or negative. Then, given that you're looking for positive solution, think about the possibilities: (+)(+)(+) = +,
++- = - , -+- = +, --- = -. Therefore, your solution is $-1 < x < 0$ or $x > 5$.
so i know how he got the $x>5$ but i don't get the $-1 < x < 0$ cuz it suppose to be $x > -1$ and what these possibilities have to do with the solution ? please explain to me in details.
(Q.2) solve for x in $x^\frac{5}{3} - 6x = x^\frac{4}{3}$
same book. The answer says: Next, factor  out an $x$ from each term: $x(x^\frac{2}{3} - x^\frac{1}{3} - 6) = 0$. The resulting expression is similar to $y^3(y^2 - y - 6)$, which factors into $y^3(y+2)(y-3)$.Similarly, you can factor $x(x^\frac{2}{3} - x^\frac{1}{3} - 6)$ into $x(x^\frac{1}{3} + 2)(x^\frac{1}{3} - 3) = 0$
I don't get how did he factor the main equation and $x^\frac{5}{3}$ became $x^\frac{2}{3}$ and $x^\frac{4}{3}$ became $x^\frac{1}{3}$. I know how to factor like this but with numbers not fractions. And also how this expression $x(x^\frac{2}{3} - x^\frac{1}{3} - 6) = 0$ is similar to that $y^3(y^2 - y - 6)$. I'm solving for x so why he got the y into the answer now ?! (weird) 
(Q.3) simplify $\frac{8}{4^\frac{2}{3}}$ 
same book. The answers says: change it to $\frac{8}{\sqrt[3]{4^2}}$ ( i get this one ) but then he said  multiply the numerator and denominator by one more cube root of 4 . so how i can multiply $\frac{8}{\sqrt[3]{4^2}}$ to $\frac{8}{ \sqrt[3]{4}}$ and get $4$ ? isn't supposed to be $\frac{8}{\sqrt[3]{4^2}}*\frac{8}{\sqrt[3]{4^2}}$equals $4$ ?
 [/INST] multiply the numerator AND denominator by one more cube root of 4.
$$\frac{8}{4^{\frac{2}{3}}} = \frac{8}{\sqrt[3]{4^2}}\cdot \frac{\sqrt[3]{4^1}}{\sqrt[3]{4^1}} = \frac{8\cdot\sqrt[3]4}{\sqrt[3]{4^3}}=\frac{8\cdot\sqrt[3]4}{4}=2\cdot\sqrt[3]4$$
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