Getting confused with rules of sqrts and negatives. So I have to simplify: $-\sqrt{(2x + 1)^6}$
What I was able to simplify that down to is $-(2x+1)^3$, however, I think I'm simplifying the inside of the square root wrong. And if I was right, what happens with the negative when I foil it out 3 times? Will it be added in the end?
 A: Restricting to $\mathbb{R}$?
$$-\sqrt{(2x+1)^6}=-|(2x+1)^3|=-|2x+1|^3$$
A: $x \in \mathbb{R}$ 
$-\sqrt{(2x + 1)^6} = -|2x + 1|^3$
$x \geq -0.5$
$-\sqrt{(2x + 1)^6} = -(2x + 1)^3$
The negative sign is outside the absolute value and outside the parenthesis.
So the result is negative for any $x$ except for $-0.5$, if that is what wonder about.
John, with my answer i was trying to explain for what values your solution are correct.
A: $-\sqrt{whatever}$ requires $whatever \ge 0$ so we must first determine which values of $x$ are a allowed so that we can be assured $(2x+1)^6 \ge 0$.
As $6$ is even $whichever^6 \ge 0$ so $(2x + 1)^6 \ge 0$ for all $x$ and $-\sqrt{(2x+1)^6}$ is defined for all $x$.
Now $\sqrt{whatsoever^{2n}} = |whatsoever^n| = |whatsoever|^n$.
[BTW.  If $n$ is odd.  And $whatsoever \ge 0$, $whatsoever^n \ge 0$ and $|whatsoever|^n = whatsoever^n$.  If $whatsoever < 0$ then $whatsoever^n < 0$ so $|whatsoever|^n = -whatsoever^n > 0$.]
So $-\sqrt{(2x + 1)^6} = -|2x + 1|^3$.
[And $-|2x + 1|^3 = -(2x + 1)^3 \le 0$ if $2x + 1 \ge 0$, and  $-|2x + 1|^3 = (2x + 1)^3 < 0$ if $2x + 1 < 0$.]
