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?

I am sorry, the thing below isnt formatted properly so i cant see what it is. Also, i don not seem to understand how the formula helps/ — QuIcKmAtHs, Feb 24 '18 at 08:36

score 1 · Accepted Answer · answered Feb 24 '18 at 08:35

1

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

answered Feb 24 '18 at 08:35

Robert Z

145,942

Thank you! Now I understand – QuIcKmAtHs Feb 24 '18 at 08:36
Nothing to add :-) – Jyrki Lahtonen Feb 24 '18 at 08:37
I got $\sqrt[3]4-1$, is that right? – QuIcKmAtHs Feb 24 '18 at 08:40
Yes, that's correct! – Robert Z Feb 24 '18 at 08:45

score 1 · Answer 2 · answered Feb 24 '18 at 08:35

1

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

answered Feb 24 '18 at 08:35

Qurultay

5,224

Sum of reciprocal of sum of 3 cube roots

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