1. ## Factor & Simplify

I'm confused on how you solve this problem D;
I don't know how to get the answer into a fraction without negative exponents.

Help?

$-4x (3x - 2)^-4/3 + 4(3x - 2)^-1/3$
-4x (3x - 2)^-4/3 + 4(3x - 2)^-1/3

A step by step explanation would really help me to understand it better.
Thank you!

2. Hello, meiyukichan!

There are several approaches to this problem.
. . Here is one of them . . .

Simplify: . $-4x (3x - 2)^{-\frac{4}{3}} + 4(3x - 2)^{-\frac{1}{3}}$
Get rid of the negative exponents now . . .

We have: . $\frac{-4x}{(3x-2)^{\frac{4}{3}}} + \frac{4}{(3x-2)^{\frac{1}{3}}}$

Get a common denominator:

. . $\frac{-4x}{(3x-2)^{\frac{4}{3}}} + \frac{4}{(3x-2)^{\frac{1}{3}}}\cdot{\color{blue}\frac{3x-2}{3x-2}} \;\;=\;\; \frac{-4}{(3x-2)^{\frac{3}{2}}} + \frac{4(3x-2)}{(3x-2)^{\frac{3}{2}}}$

. . $= \;\frac{-4x+4(3x-2)}{(3x-2)^{\frac{3}{2}}} \;\;=\;\;\frac{-4x + 12x - 8}{(3x-2)^{\frac{3}{2}}} \;\;=\;\;\boxed{\frac{8x-8}{(3x-2)^{\frac{3}{2}}}}$

3. Originally Posted by Soroban

Thank you, thank you, thank you!
Very helpful and now I understand that with negative exponents, you're suppose to write the inverse of it to get rid of the negative... right? heh.