Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$

Put $a = \sqrt[3]{5x+7}, b = \sqrt[3]{5x-12}\implies a - b = 1, a^3 - b^3 =19\implies a^2+ab+b^2 = \dfrac{a^3-b^3}{a-b} = \dfrac{19}{1} = 19\implies (a-b)^2+3ab = 19\implies 1^2+3ab = 19 \implies ab = 6\implies (b+1)b = 6\implies b^2+b-6 = 0\implies (b+3)(b-2)=0\implies b = 2, -3\implies 5x-12 = 2^3, (-3)^3 = 8,-27\implies 5x = 20, -15\implies x = 4, -3.$

Before cubing, separate your roots:

$$\sqrt[3]{5x+7}=\sqrt[3]{5x-12}+1$$

Now cube:

$$5x + 7 = 5x - 12 + 3\sqrt[3]{5x-12}^2 + 3\sqrt[3]{5x-12} + 1$$

Simplify:

$$ \sqrt[3]{5x-12}^2 + \sqrt[3]{5x-12} - 6 = 0$$

Solve for $\sqrt[3]{5x-12}$ (it is a quadratic equation). Finally solve to $x$.

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 $x=-3$.