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.

Actually, there is a nice, generic solution to equations of this type, namely equations in the form
$$\sqrt[3]{f(x)}+\sqrt[3]{g(x)} = C,$$
especially if $f(x)$ and $g(x)$ sum to a constant.
Cubing and rearranging, we get
$$f(x) + g(x) + 3C\sqrt[3]{f(x)g(x)}=C^3,$$
or
$$\sqrt[3]{f(x)g(x)} = \frac{C^3-f(x)-g(x)}{3C}.$$
Cubing again, we get
$$f(x)g(x) = \frac{(C^3-f(x)-g(x))^3}{27C^3}.$$
If $f(x)$ and $g(x)$ are linear in $x$, this is a cubic equation. But if $f(x)+g(x)=A$ const., we get
$$f(x)[A-f(x)] = \frac{(C^3-A)^3}{27C^3},$$
or
$$f(x)^2 - Af(x) + \left[\frac{(C^3-A)}{3C}\right]^3 = 0,$$
which is quadratic in $f(x)$.
In the specific example, $f(x)=5x+7$, $g(x)=-5x+12$, $A=f(x)+g(x)=19$, $C=1$, and we get
$$f(x)^2 - 19f(x) - 216=0,$$
which yields $f(x)=5x+7=27$ or $f(x)=5x+7=-8$, i.e., $x=4$ or $x=-3$.
