
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
 LinkBack  Thread Tools  Display Modes 
May 2nd, 2018, 01:08 AM  #1 
Newbie Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0  Irrational equations with cube roots
$\begin{aligned}&\sqrt[3]{x+1}+\sqrt[3]{3x+1}=\sqrt[3]{x1}\\ &\Leftrightarrow(x+1)+(3x+1)+3\sqrt[3]{(x+1)(3x+1)}(\sqrt[3]{x+1}+\sqrt[3]{3x+1})=x1\\ &\Rightarrow 3(x+1)+3\sqrt[3]{(x+1)(3x+1)}\cdot\sqrt[3]{x1}=0\ \ldots\ x=1\end{aligned}$ The part that confuses me is the implication. How did we go from $\sqrt[3]{x+1}+\sqrt[3]{3x+1}$ to $\sqrt[3]{x1}$ Any help much appreciated! 
May 2nd, 2018, 01:15 AM  #2 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry 
It's stated in the first line, so basically you just substituted the first line to the second and you get the third.


Tags 
cube, equations, irrational, roots 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
cube roots of unity  not3bad  Complex Analysis  2  December 20th, 2014 02:35 AM 
Cube roots of 1  agentredlum  Algebra  24  December 31st, 2013 03:57 AM 
Derivative of cube roots and trig  imcutenfresa  Calculus  1  September 20th, 2009 05:34 AM 
Three cube roots of 1  dequinox  Complex Analysis  1  June 9th, 2009 03:00 PM 