5
$\begingroup$

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

enter image description here

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?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Shailesh, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Note: $\sqrt[3]4=\sqrt[3]2^2$ $\endgroup$ – J. W. Tanner Jun 16 at 19:41
  • $\begingroup$ Have you noticed that each term is $$\frac{1}{\cbrt{t^2}+\cbrt{t(t+1)}+\cbrt{(t+1)^2}}$$ $\endgroup$ – Rhys Hughes Jun 16 at 19:41
  • 4
    $\begingroup$ @RhysHughes the command to get cubic root is \sqrt[3]{} $\endgroup$ – Adam Latosiński Jun 16 at 19:45
  • 1
    $\begingroup$ Possible duplicate of Sum of reciprocal of sum of 3 cube roots $\endgroup$ – YuiTo Cheng Jun 17 at 7:12
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $4-3$. $\endgroup$ – Adam Latosiński Jun 16 at 19:49
  • $\begingroup$ @AdamLatosiński thank you. It was a silly mistake, I edited my answer $\endgroup$ – user289143 Jun 16 at 19:54
  • $\begingroup$ This is NOT the sum! $\endgroup$ – Dr. Sonnhard Graubner Jun 16 at 19:55
  • $\begingroup$ @Dr.SonnhardGraubner my answer is the same as yours... $\endgroup$ – user289143 Jun 16 at 19:55
  • $\begingroup$ Yes your answer was updated! $\endgroup$ – Dr. Sonnhard Graubner Jun 16 at 19:56
6
$\begingroup$

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 $x=\sqrt[3]{3}$, $y=\sqrt[3]{4}$.

$\endgroup$
3
$\begingroup$

Hint to explain the quote:

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.