Sum of reciprocal of sum of 3 cube roots

Question

Simplify $$\frac{1}{\sqrt[3]1+\sqrt[3]2+\sqrt[3]4}+\frac{1}{\sqrt[3]4+\sqrt[3]6+\sqrt[3]9}+\frac{1}{\sqrt[3]9+\sqrt[3]{12}+\sqrt[3]{16}}$$

I have no idea how to do this. I tried using the idea of multiplying the conjugate to every term, but I seem to be getting no where. Is there any hint as to how to approach this?

Answer 1

Hint. Notice that $$\frac{1}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}=\frac{\sqrt[3]{a}-\sqrt[3]{b}}{a-b}.$$

Answer 2

Use the $(a-b)(a^2+ab+b^2)=a^3-b^3$, then for example $$\frac{1}{\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{1-\sqrt[3]{2}}{1-2} $$