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Summing cube roots in fractions

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4

$begingroup$

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?

arithmetic

share|cite|improve this question

edited 9 hours ago

Akerbeltz

9221426

asked 9 hours ago

JamminermitJamminermit

12310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: $sqrt[3]4=sqrt[3]2^2$
  $endgroup$
  – J. W. Tanner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you noticed that each term is $$frac1cbrtt^2+cbrtt(t+1)+cbrt(t+1)^2$$
  $endgroup$
  – Rhys Hughes
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @RhysHughes the command to get cubic root is sqrt[3]
  $endgroup$
  – Adam Latosiński
  9 hours ago
  
  

  
  
  
  

add a comment | 

4

$begingroup$

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?

arithmetic

share|cite|improve this question

edited 9 hours ago

Akerbeltz

9221426

asked 9 hours ago

JamminermitJamminermit

12310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: $sqrt[3]4=sqrt[3]2^2$
  $endgroup$
  – J. W. Tanner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you noticed that each term is $$frac1cbrtt^2+cbrtt(t+1)+cbrt(t+1)^2$$
  $endgroup$
  – Rhys Hughes
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @RhysHughes the command to get cubic root is sqrt[3]
  $endgroup$
  – Adam Latosiński
  9 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

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?

arithmetic

share|cite|improve this question

edited 9 hours ago

Akerbeltz

9221426

asked 9 hours ago

JamminermitJamminermit

12310

$endgroup$

share|cite|improve this question

edited 9 hours ago

Akerbeltz

9221426

asked 9 hours ago

JamminermitJamminermit

12310

share|cite|improve this question

edited 9 hours ago

Akerbeltz

9221426

asked 9 hours ago

JamminermitJamminermit

12310

edited 9 hours ago

Akerbeltz

9221426

edited 9 hours ago

Akerbeltz

9221426

asked 9 hours ago

JamminermitJamminermit

12310

asked 9 hours ago

JamminermitJamminermit

12310

3 Answers
3

active

oldest

votes

5

$begingroup$

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
$$ frac1x^2+xy+y^2 = fracx-yx^3-y^3$$
If you use that for $x=1$, $y=sqrt[3]2$, you get
$$ frac11+sqrt[3]2+sqrt[3]4 = fracsqrt[3]2-12-1 = sqrt[3]2-1$$
To get the other two fractions use $x=sqrt[3]2$, $y=sqrt[3]3$ and $x=sqrt[3]3$, $y=sqrt[3]4$.

share|cite|improve this answer

answered 9 hours ago

Adam LatosińskiAdam Latosiński

4,397416

$endgroup$

add a comment | 

4

$begingroup$

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 $fracsqrt[3]2-sqrt[3]1sqrt[3]2-sqrt[3]1$, the second one by $fracsqrt[3]3-sqrt[3]2sqrt[3]3-sqrt[3]2$ and the third one by $fracsqrt[3]4-sqrt[3]3sqrt[3]4-sqrt[3]3$ and you get
$$fracsqrt[3]2-sqrt[3]12-1+fracsqrt[3]3-sqrt[3]23-2+fracsqrt[3]4-sqrt[3]34-3=sqrt[3]2-1+sqrt[3]3-sqrt[3]2+sqrt[3]4-sqrt[3]3=sqrt[3]4-1$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user289143user289143

1,575314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're wrong: $$9-4 = (sqrt[3]9-sqrt[3]4)(sqrt[3]9^2+sqrt[3]9cdot 4+sqrt[3]4^2)$$ and similarily for $16-9$. What you need to decompose in this way is $3-2$ and $4-3$.
  $endgroup$
  – Adam Latosiński
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AdamLatosiński thank you. It was a silly mistake, I edited my answer
  $endgroup$
  – user289143
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is NOT the sum!
  $endgroup$
  – Dr. Sonnhard Graubner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner my answer is the same as yours...
  $endgroup$
  – user289143
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes your answer was updated!
  $endgroup$
  – Dr. Sonnhard Graubner
  9 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Hint to explain the quote:

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

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

8,2621723

$endgroup$

add a comment | 

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5

$begingroup$

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
$$ frac1x^2+xy+y^2 = fracx-yx^3-y^3$$
If you use that for $x=1$, $y=sqrt[3]2$, you get
$$ frac11+sqrt[3]2+sqrt[3]4 = fracsqrt[3]2-12-1 = sqrt[3]2-1$$
To get the other two fractions use $x=sqrt[3]2$, $y=sqrt[3]3$ and $x=sqrt[3]3$, $y=sqrt[3]4$.

share|cite|improve this answer

answered 9 hours ago

Adam LatosińskiAdam Latosiński

4,397416

$endgroup$

add a comment | 

5

$begingroup$

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
$$ frac1x^2+xy+y^2 = fracx-yx^3-y^3$$
If you use that for $x=1$, $y=sqrt[3]2$, you get
$$ frac11+sqrt[3]2+sqrt[3]4 = fracsqrt[3]2-12-1 = sqrt[3]2-1$$
To get the other two fractions use $x=sqrt[3]2$, $y=sqrt[3]3$ and $x=sqrt[3]3$, $y=sqrt[3]4$.

share|cite|improve this answer

answered 9 hours ago

Adam LatosińskiAdam Latosiński

4,397416

$endgroup$

add a comment | 

$begingroup$

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
$$ frac1x^2+xy+y^2 = fracx-yx^3-y^3$$
If you use that for $x=1$, $y=sqrt[3]2$, you get
$$ frac11+sqrt[3]2+sqrt[3]4 = fracsqrt[3]2-12-1 = sqrt[3]2-1$$
To get the other two fractions use $x=sqrt[3]2$, $y=sqrt[3]3$ and $x=sqrt[3]3$, $y=sqrt[3]4$.

share|cite|improve this answer

answered 9 hours ago

Adam LatosińskiAdam Latosiński

4,397416

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Adam LatosińskiAdam Latosiński

4,397416

answered 9 hours ago

Adam LatosińskiAdam Latosiński

4,397416

answered 9 hours ago

Adam LatosińskiAdam Latosiński

4,397416

4

$begingroup$

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 $fracsqrt[3]2-sqrt[3]1sqrt[3]2-sqrt[3]1$, the second one by $fracsqrt[3]3-sqrt[3]2sqrt[3]3-sqrt[3]2$ and the third one by $fracsqrt[3]4-sqrt[3]3sqrt[3]4-sqrt[3]3$ and you get
$$fracsqrt[3]2-sqrt[3]12-1+fracsqrt[3]3-sqrt[3]23-2+fracsqrt[3]4-sqrt[3]34-3=sqrt[3]2-1+sqrt[3]3-sqrt[3]2+sqrt[3]4-sqrt[3]3=sqrt[3]4-1$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user289143user289143

1,575314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're wrong: $$9-4 = (sqrt[3]9-sqrt[3]4)(sqrt[3]9^2+sqrt[3]9cdot 4+sqrt[3]4^2)$$ and similarily for $16-9$. What you need to decompose in this way is $3-2$ and $4-3$.
  $endgroup$
  – Adam Latosiński
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AdamLatosiński thank you. It was a silly mistake, I edited my answer
  $endgroup$
  – user289143
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is NOT the sum!
  $endgroup$
  – Dr. Sonnhard Graubner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner my answer is the same as yours...
  $endgroup$
  – user289143
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes your answer was updated!
  $endgroup$
  – Dr. Sonnhard Graubner
  9 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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 $fracsqrt[3]2-sqrt[3]1sqrt[3]2-sqrt[3]1$, the second one by $fracsqrt[3]3-sqrt[3]2sqrt[3]3-sqrt[3]2$ and the third one by $fracsqrt[3]4-sqrt[3]3sqrt[3]4-sqrt[3]3$ and you get
$$fracsqrt[3]2-sqrt[3]12-1+fracsqrt[3]3-sqrt[3]23-2+fracsqrt[3]4-sqrt[3]34-3=sqrt[3]2-1+sqrt[3]3-sqrt[3]2+sqrt[3]4-sqrt[3]3=sqrt[3]4-1$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user289143user289143

1,575314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're wrong: $$9-4 = (sqrt[3]9-sqrt[3]4)(sqrt[3]9^2+sqrt[3]9cdot 4+sqrt[3]4^2)$$ and similarily for $16-9$. What you need to decompose in this way is $3-2$ and $4-3$.
  $endgroup$
  – Adam Latosiński
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AdamLatosiński thank you. It was a silly mistake, I edited my answer
  $endgroup$
  – user289143
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is NOT the sum!
  $endgroup$
  – Dr. Sonnhard Graubner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner my answer is the same as yours...
  $endgroup$
  – user289143
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes your answer was updated!
  $endgroup$
  – Dr. Sonnhard Graubner
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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 $fracsqrt[3]2-sqrt[3]1sqrt[3]2-sqrt[3]1$, the second one by $fracsqrt[3]3-sqrt[3]2sqrt[3]3-sqrt[3]2$ and the third one by $fracsqrt[3]4-sqrt[3]3sqrt[3]4-sqrt[3]3$ and you get
$$fracsqrt[3]2-sqrt[3]12-1+fracsqrt[3]3-sqrt[3]23-2+fracsqrt[3]4-sqrt[3]34-3=sqrt[3]2-1+sqrt[3]3-sqrt[3]2+sqrt[3]4-sqrt[3]3=sqrt[3]4-1$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user289143user289143

1,575314

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user289143user289143

1,575314

answered 9 hours ago

user289143user289143

1,575314

answered 9 hours ago

user289143user289143

1,575314

3

$begingroup$

Hint to explain the quote:

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

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

8,2621723

$endgroup$

add a comment | 

3

$begingroup$

Hint to explain the quote:

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

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

8,2621723

$endgroup$

add a comment | 

$begingroup$

Hint to explain the quote:

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

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

8,2621723

$endgroup$

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

8,2621723

answered 9 hours ago

J. W. TannerJ. W. Tanner

8,2621723

answered 9 hours ago

J. W. TannerJ. W. Tanner

8,2621723