<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] Following up on OP's "almost-there" attempt (and correcting the $-1\color{red}{2}\,$ in the second equation):

$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1 \tag{1}$
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\color{red}{2}}-\sqrt[3]{5x+7})=-6 $

The rightmost factor is $-1$ per the original equation $(1)\,$, which leaves:
$$-\sqrt[3]{(5x-12)(5x+7)}=-6 \tag{2}$$
Then the numbers $\,\sqrt[3]{5x+7}\,$ and $\,-\sqrt[3]{5x-12}\,$ have sum $\,1\,$ per $(1)\,$, and product $\,-6\,$ per $(2)\,$, so they are the roots of the quadratic $\,t^2-t-6=0 \iff t \in \{-2, 3\}\,$:

*

*$\sqrt[3]{5x+7} = -2 \iff 5x +7 = -8 \iff x = -3$


*$\sqrt[3]{5x+7} = 3 \iff 5x + 7 = 27 \iff x = 4$
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