Solve an equation with cube roots How can I solve the following equation:
$$\frac{(34-x)\sqrt[3]{x+1}-(x+1)\sqrt[3]{34-x}}{\sqrt[3]{34-x}-\sqrt[3]{x+1}}=30$$
Attempts:
By substituting $\sqrt[3]{34-x}=a$ and $\sqrt[3]{x+1}=b$ we get:
$$\frac{a^3b-b^3a}{a-b}=\frac{ab(a^2-b^2)}{a-b} = ab(a-b)=30$$
Although it looks good, it doesn't lead to anything useful for me.

Let $\sqrt[3]{34-x}=a$ and $\sqrt[3]{x+1}=b$.
Thus, $$a^3+b^3=35$$ and $$ab(a+b)=30,$$ which gives
$$(a+b)^3=125$$ or $$a+b=5.$$
Can you end it now?
I got $$\{26,7\}$$
