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May 2nd, 2018, 12:08 AM   #1
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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!
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May 2nd, 2018, 12:15 AM   #2
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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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