**1 :**Simplify the type one radical:

First we should observe that the given radical really is a type one radical as each of the factors in the radicand is being multiplied against each of the other factors. To simplify the radical we want to group each of the (multiplicative) terms of the radicand into “perfect cube” pieces (since the root-value is ), and then the leftover “remainder” piece (just like we did with the prime factors in the numeric radical instance.)

The good news is that the simplification *process* is the same as the numeric examples we’ve seen (albeit with a looser definition
of ‘prime factor’). It should not be a total surprise that the factored form treats each (multiplicative) factor as a ‘prime
factor’ in terms of simplifying the radical, similar to the how we simplify numeric radicals. Indeed, we’ve already
seen that roots are analogous to ‘prime-numbers’ when it comes to decomposing polynomials to their ‘most basic
parts’, so it shouldn’t be surprising that they end up being treated similarly in other situations (like radicals) as
well!

Unfortunately, the bad news is that the *result* isn’t quite as straightforward as in the case of the numeric radical case. To
motivate the next part, we will first proceed with an example.

**1.1 :**Find all that satisfy the equation

This example seems straight forward. According to our rules of simplifying it seems like . Indeed, if we do that then we get as an answer and, plugging that back in to the original equation, we get . So far so good right? But what about ? If we plug that in we will also get , so is is irrelevantalso a solutionworks, but isn’t a solution since square roots are only positive . So, why didn’t our method give us this answer as well?

If we look closely at the previous example it may become apparent that the reason that is also a solution, is that the in the
original equation kills the negative sign. Thus, since is a solution, and the even power nullifies the effect of the negative sign, then
is also a solution (that is to say; since and , there is no difference between the two once we square them). But
understanding *why* the solution is valid is not quite the same thing as figuring out why we didn’t find it the first time.
After all if our methodology were correct, it should have given us *all* the valid solutions the first time around
right?

As it turns out, this all comes down to the fact that the radical (symbol) is defined to *only give positive outputs*. To
understand why this is the case though, we need to revisit the primary reason we have radicals which is what we discuss
next.