Skip to main content
9 events
when toggle format what by license comment
Apr 13, 2017 at 12:19 history edited CommunityBot
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
May 6, 2013 at 21:50 answer added Dave L. Renfro timeline score: 8
May 6, 2013 at 2:32 vote accept Samuel Tan
May 6, 2013 at 1:56 comment added Julian Rosen In general, for $(ax+b)^{1/n}$, you can multiply numerator and denominator by $\displaystyle \sum_{j=0}^{n-1} (a(x+h)+b)^{\frac{j}{n}}(ax+b)^{\frac{n-1-j}{n}}$. After multiplying out, the numerator will just be $ah$.
May 6, 2013 at 1:51 comment added Samuel Tan How would this work for $(ax+b)^{1/3}$?
May 6, 2013 at 1:50 answer added user17762 timeline score: 4
May 6, 2013 at 1:50 comment added Samuel Tan Right...why didn't I see that before..duh! Thank you!
May 6, 2013 at 1:48 comment added Julian Rosen Hint: Multiple the numerator and denominator by $\sqrt{a(x+h)+b}+\sqrt{ax+b}$.
May 6, 2013 at 1:47 history asked Samuel Tan CC BY-SA 3.0