<s>[INST] Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$ I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this point I'm out of stuff to do. Any help would be appreciated. Thanks in advance.
 [/INST] Let's see. We have
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$$
now set $x=y+\frac12$. Then
$$\sqrt[3]{5y+\tfrac{19}2}-\sqrt[3]{5y-\tfrac{19}2}=1$$
Now cubing yields
$$19-3\sqrt[3]{(5y)^2-(\tfrac{19}2)^2}\left(\sqrt[3]{5y+\tfrac{19}2}-\sqrt[3]{5y-\tfrac{19}2}\right)=1$$
which rewrites to
$$\sqrt[3]{(5y)^2-(\tfrac{19}2)^2}=6$$
Which is easy to solve, $25y^2=6^3+(\frac{19}2)^2$ or $y=\pm \frac 72$. Now we conclude $x=4$ or $x=-3$.
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