# Summing cube roots in fractions [closed]

I found this problem, and understand the solution, but do not understand why they made the first assumption. The problem:

The first line of the solution says that:

The cube root of $$1$$ plus the cube root of $$2$$ plus the cube root of $$4$$ is a factor of $$2-1$$.

Why are you meant to assume this to solve the problem?

## closed as off-topic by José Carlos Santos, Shailesh, Cesareo, YuiTo Cheng, Thomas ShelbyJun 17 at 11:31

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• Note: $\sqrt[3]4=\sqrt[3]2^2$ – J. W. Tanner Jun 16 at 19:41
• Have you noticed that each term is $$\frac{1}{\cbrt{t^2}+\cbrt{t(t+1)}+\cbrt{(t+1)^2}}$$ – Rhys Hughes Jun 16 at 19:41
• @RhysHughes the command to get cubic root is \sqrt[3]{} – Adam Latosiński Jun 16 at 19:45
• Possible duplicate of Sum of reciprocal of sum of 3 cube roots – YuiTo Cheng Jun 17 at 7:12

Probably they are telling you a way to rationalize the denominator so you can do the sum.

$$1=2-1=(\sqrt[3]{2}-\sqrt[3]{1})(\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1})$$

Similarly

$$1=3-2=(\sqrt[3]{3}-\sqrt[3]{2})(\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4})$$

and

$$1=4-3=(\sqrt[3]{4}-\sqrt[3]{3})(\sqrt[3]{16}+\sqrt[3]{12}+\sqrt[3]{9})$$

so you can multiply the first fraction by $$\frac{\sqrt[3]{2}-\sqrt[3]{1}}{\sqrt[3]{2}-\sqrt[3]{1}}$$, the second one by $$\frac{\sqrt[3]{3}-\sqrt[3]{2}}{\sqrt[3]{3}-\sqrt[3]{2}}$$ and the third one by $$\frac{\sqrt[3]{4}-\sqrt[3]{3}}{\sqrt[3]{4}-\sqrt[3]{3}}$$ and you get $$\frac{\sqrt[3]{2}-\sqrt[3]{1}}{2-1}+\frac{\sqrt[3]{3}-\sqrt[3]{2}}{3-2}+\frac{\sqrt[3]{4}-\sqrt[3]{3}}{4-3}=\sqrt[3]{2}-1+\sqrt[3]{3}-\sqrt[3]{2}+\sqrt[3]{4}-\sqrt[3]{3}=\sqrt[3]{4}-1$$

• You're wrong: $$9-4 = (\sqrt[3]{9}-\sqrt[3]{4})(\sqrt[3]{9^2}+\sqrt[3]{9\cdot 4}+\sqrt[3]{4^2})$$ and similarily for $16-9$. What you need to decompose in this way is $3-2$ and $4-3$. – Adam Latosiński Jun 16 at 19:49
• @AdamLatosiński thank you. It was a silly mistake, I edited my answer – user289143 Jun 16 at 19:54
• This is NOT the sum! – Dr. Sonnhard Graubner Jun 16 at 19:55
• @Dr.SonnhardGraubner my answer is the same as yours... – user289143 Jun 16 at 19:55
• Yes your answer was updated! – Dr. Sonnhard Graubner Jun 16 at 19:56

They are pointing to the fact (it's not an assumption) that $$(x^2+xy+y^2)(x-y) = x^3-y^3$$ that is $$\frac{1}{x^2+xy+y^2} = \frac{x-y}{x^3-y^3}$$ If you use that for $$x=1$$, $$y=\sqrt[3]{2}$$, you get $$\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}} = \frac{\sqrt[3]{2}-1}{2-1} = \sqrt[3]{2}-1$$ To get the other two fractions use $$x=\sqrt[3]{2}$$, $$y=\sqrt[3]{3}$$ and $$x=\sqrt[3]{3}$$, $$y=\sqrt[3]{4}$$.

Hint to explain the quote:

use $$x^3-1=(x-1)(1+x+x^2)$$ with $$x=\sqrt[3]2.$$