# Simplifying question

• Mar 3rd 2009, 07:39 AM
Barbor
Simplifying question
Hello,

I was wondering how to simplify the following expression:

http://img99.imageshack.us/img99/7948/93261763.jpg

I hope the picture is clear enough and that someone can help me :)
• Mar 3rd 2009, 08:06 AM
lanzailan
Hello Barbor..
let me try....

(1/6x)^(1/3).(4/6)^2/3
=(1/6)^(1/3).(4/6)^(2/3).x^(1/3).x^(2/3)
=(1/6)^(1/3).(4/6)^2(1/3).x^(1/3+2/3)
=(1/6)^(1/3).(4/6)^(1/3).(4/6)^(1/3).x
=(1/6)^(1/3).(4)^(1/3).(1/6)^(1/3).(4)^(1/3).(1/6)^(1/3).x
=(1/6)^(1/3+1/3+1/3).4^(1/3).x
=(1/6)^(3/3).4^(1/3).x
=1/6.4^(1/3).x
• Mar 3rd 2009, 11:59 AM
Soroban
Hello, Barbor!

Quote:

$\displaystyle \left(\frac{1}{6}\,x\right)^{\frac{1}{3}}\left(\fr ac{4}{6}\,x\right)^{\frac{2}{3}}$

We have: .$\displaystyle \left(\frac{1}{6}\right)^{\frac{1}{3}}\!\!\cdot x^{\frac{1}{3}} \cdot\!\left(\frac{4}{6}\right)^{\frac{2}{3}}\!\!\ cdot x^{\frac{2}{3}} \;\;=\;\;\frac{1^{\frac{1}{3}}}{6^{\frac{1}{3}}} \cdot\frac{4^{\frac{2}{3}}}{6^{\frac{2}{3}}} \cdot x^{\frac{1}{3}} \cdot x^{\frac{2}{3}} \;\;=\; \;\frac{4^{\frac{2}{3}}}{6^1} \cdot x^1$

. . . $\displaystyle =\;\;\frac{(2^2)^{\frac{2}{3}}}{6}\,x \;\;=\;\; \frac{2^{\frac{4}{3}}}{2\cdot3}\,x \;\;=\;\; \frac{2^{\frac{1}{3}}}{3}\,x \;=\;\frac{\sqrt[3]{2}}{3}\,x$

• Mar 3rd 2009, 02:04 PM
Barbor
Thank you very much!