
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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May 2nd, 2018, 12:08 AM  #1 
Newbie Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0  Irrational equations with cube roots
$\begin{aligned}&\sqrt[3]{x+1}+\sqrt[3]{3x+1}=\sqrt[3]{x1}\\ &\Leftrightarrow(x+1)+(3x+1)+3\sqrt[3]{(x+1)(3x+1)}(\sqrt[3]{x+1}+\sqrt[3]{3x+1})=x1\\ &\Rightarrow 3(x+1)+3\sqrt[3]{(x+1)(3x+1)}\cdot\sqrt[3]{x1}=0\ \ldots\ x=1\end{aligned}$ 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, 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.


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cube, equations, irrational, roots 
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