Q:
$\displaystyle \frac {(1 + 2x)^{ \frac {3}{2}} - 3 \sqrt {1 + 2x}}{6}$
$\displaystyle = \frac { \sqrt {1 + 2x} [(1 + 2x) - 3]}{6}$
$\displaystyle = \frac { \sqrt {1 + 2x} (2x - 2)}{6}$
$\displaystyle = \frac { \sqrt {1 + 2x} (x - 1) \cdot 2}{6}$
$\displaystyle = \frac {(x - 1) \sqrt {1 + 2x}}{3}$
Hey thanks Jhevon! So any time you do try to solve for "x" and it works out you take that du value which in this case was $\displaystyle \frac{1}{2}$ and simplify it along with the rest of the equation? I guess what I am trying to ask is... you DO NOT MOVE THE DU OUT IN FRONT, that would be bad wouldn't it? Thanks again for all of the tutoring tonight...
Thanks!
-qbkr21