For some reason, I keep forgetting the root rules for the GRE. This is what I have been able to find on the internet and would like some confirmation.
If the problem has x^2 = 9, then x could be -3 or 3.
If the problem has \sqrt{x^2}=9, then x is only 3.
If the problem has \sqrt[3]{x^3}=8, then x is only 2.
If the problem has x^3=8, then x could be -2 or 2.
Basically, if the user has to introduce a root, the answer could be the negative. If the radical or root already exists, then you have to consider the degree of the value you are taking. The \sqrt{x^2} is the absolute value of x. But if it is the cube root of x^3, then the answer is the positive root.
There’s a mistake for the fourth expression. The only solution to x^3 = 8 is x=2; if you do (-2) \cdot (-2) \cdot (-2) it will result in -8.
Personally, the only thing I remember is that the only exception is your second rule, i.e. \sqrt{x^2} = 9 \rightarrow x=3 \text{ only}. More generally I just remember to take only the positive when the root is even.
Because as you can see, every other case is consistent with the general math convention
