My Math Forum Irrational equations with cube roots

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 May 2nd, 2018, 12:08 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]{x-1}\\ &\Leftrightarrow(x+1)+(3x+1)+3\sqrt[3]{(x+1)(3x+1)}(\sqrt[3]{x+1}+\sqrt[3]{3x+1})=x-1\\ &\Rightarrow 3(x+1)+3\sqrt[3]{(x+1)(3x+1)}\cdot\sqrt[3]{x-1}=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 $\sqrt[3]{x-1}$ Any help much appreciated!
 May 2nd, 2018, 12:15 AM #2 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,000 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. Thanks from bongcloud

 Tags cube, equations, irrational, roots

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post not3bad Complex Analysis 2 December 20th, 2014 01:35 AM agentredlum Algebra 24 December 31st, 2013 02:57 AM imcutenfresa Calculus 1 September 20th, 2009 04:34 AM dequinox Complex Analysis 1 June 9th, 2009 02:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top