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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]{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: 2,001 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. Thanks from bongcloud

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