Solve an equation with cube roots

Question

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.

Answer 1

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 $$\{26,7\}$$

Answer 2

if $x=7$ then $x+1 = 8$ has an integer cube root and $34-x = 27$ has an integer cube root. Also, you get an answer to the original problem.

worth checking what happens when $x=26...$

Yes, $x=26$ also works.