Find all real number(s) $x$ satisfying the equation $(x +1)^3$ = $x^3$ , where $y$ denotes the fractional part of $y$

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Find all real number(s) $x$ satisfying the equation $(x +1)^3$ = $x^3$ , where $y$ denotes the fractional part of $y$ , for example $3.1416ldots=0.1416ldots$.

I am trying all positive real numbers from $1,2,dots$ but I didn't get any decimals.

Is there a smarter way to solve this problem? ... Please advise.

real-numbers integers decimal-expansion

share|cite|improve this question

edited Jul 18 at 11:46

Asaf Karagila♦

292k31403733

asked Jul 18 at 9:45

Math Tise

3727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use $ x$ for $ x$.
  – Shaun
  Jul 18 at 9:47
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You're trying all positive real numbers? xD
  – principal-ideal-domain
  Jul 18 at 9:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To begn with, the left side is a number between $0$ and $1$ , so the right side must be the same. This tells you that all solutions must fall in the range $[0,1)$ at best, if there are any solutions.
  – MPW
  Jul 18 at 9:55
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Find all real number(s) $x$ satisfying the equation $(x +1)^3$ = $x^3$ , where $y$ denotes the fractional part of $y$ , for example $3.1416ldots=0.1416ldots$.

I am trying all positive real numbers from $1,2,dots$ but I didn't get any decimals.

Is there a smarter way to solve this problem? ... Please advise.

real-numbers integers decimal-expansion

share|cite|improve this question

edited Jul 18 at 11:46

Asaf Karagila♦

292k31403733

asked Jul 18 at 9:45

Math Tise

3727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use $ x$ for $ x$.
  – Shaun
  Jul 18 at 9:47
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You're trying all positive real numbers? xD
  – principal-ideal-domain
  Jul 18 at 9:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To begn with, the left side is a number between $0$ and $1$ , so the right side must be the same. This tells you that all solutions must fall in the range $[0,1)$ at best, if there are any solutions.
  – MPW
  Jul 18 at 9:55
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Find all real number(s) $x$ satisfying the equation $(x +1)^3$ = $x^3$ , where $y$ denotes the fractional part of $y$ , for example $3.1416ldots=0.1416ldots$.

I am trying all positive real numbers from $1,2,dots$ but I didn't get any decimals.

Is there a smarter way to solve this problem? ... Please advise.

real-numbers integers decimal-expansion

share|cite|improve this question

edited Jul 18 at 11:46

Asaf Karagila♦

292k31403733

asked Jul 18 at 9:45

Math Tise

3727

Find all real number(s) $x$ satisfying the equation $(x +1)^3$ = $x^3$ , where $y$ denotes the fractional part of $y$ , for example $3.1416ldots=0.1416ldots$.

I am trying all positive real numbers from $1,2,dots$ but I didn't get any decimals.

Is there a smarter way to solve this problem? ... Please advise.

real-numbers integers decimal-expansion

share|cite|improve this question

edited Jul 18 at 11:46

Asaf Karagila♦

292k31403733

asked Jul 18 at 9:45

Math Tise

3727

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 11:46

Asaf Karagila♦

292k31403733

edited Jul 18 at 11:46

Asaf Karagila♦

292k31403733

edited Jul 18 at 11:46

Asaf Karagila♦

292k31403733

292k31403733

asked Jul 18 at 9:45

Math Tise

3727

asked Jul 18 at 9:45

Math Tise

3727

asked Jul 18 at 9:45

Math Tise

3727

3727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use $ x$ for $ x$.
  – Shaun
  Jul 18 at 9:47
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You're trying all positive real numbers? xD
  – principal-ideal-domain
  Jul 18 at 9:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To begn with, the left side is a number between $0$ and $1$ , so the right side must be the same. This tells you that all solutions must fall in the range $[0,1)$ at best, if there are any solutions.
  – MPW
  Jul 18 at 9:55
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use $ x$ for $ x$.
  – Shaun
  Jul 18 at 9:47
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You're trying all positive real numbers? xD
  – principal-ideal-domain
  Jul 18 at 9:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To begn with, the left side is a number between $0$ and $1$ , so the right side must be the same. This tells you that all solutions must fall in the range $[0,1)$ at best, if there are any solutions.
  – MPW
  Jul 18 at 9:55
  

  
  
  
  

Use $ x$ for $ x$.
– Shaun
Jul 18 at 9:47

Use $ x$ for $ x$.
– Shaun
Jul 18 at 9:47

3

3

You're trying all positive real numbers? xD
– principal-ideal-domain
Jul 18 at 9:52

You're trying all positive real numbers? xD
– principal-ideal-domain
Jul 18 at 9:52

To begn with, the left side is a number between $0$ and $1$ , so the right side must be the same. This tells you that all solutions must fall in the range $[0,1)$ at best, if there are any solutions.
– MPW
Jul 18 at 9:55

To begn with, the left side is a number between $0$ and $1$ , so the right side must be the same. This tells you that all solutions must fall in the range $[0,1)$ at best, if there are any solutions.
– MPW
Jul 18 at 9:55

add a comment | 

3 Answers
3

active

oldest

votes

up vote
5
down vote



accepted

Fractional part is always in $[0,1)$. So the domain of $g(x)=x^3$ you are looking for is such that $R_gin[0,1)$, which happens to be $[0,1)$.

$y=(x+1)^3$ is basically $y=(x+1)^3$ chopped into appropriate pieces and translated down by some $kinmathbbZ$ where $k=lfloor(x+1)^3rfloor$

So we solve
$$(x+1)^3-k=x^3$$
which gives
$$k=3x^2+3x+1$$

Since for $[0,1)$, $0le(x+1)^3<8$, we check
$$3x^2+3x+1=0,1,2,3,4,5,6,7$$
whose positive roots give the answers.

share|cite|improve this answer

edited Jul 18 at 10:06

answered Jul 18 at 9:55

Karn Watcharasupat

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact the fractional part is always in $[0,1)$
  – MPW
  Jul 18 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MPW ugh my bad! edited. thanks a lot
  – Karn Watcharasupat
  Jul 18 at 9:57
  

  
  
  
  

add a comment | 

up vote
5
down vote

Hints:

• $0 le y lt 1$


• so any solution to $(x +1)^3 = x^3$ has $0 le x^3 lt 1$ and thus $0 le x lt 1$


• so $1 le x+1 lt 2$ and $1 le (x+1)^3 lt 8$


• any solution has $(x+1)^3 = x^3 +n$ for $n in 1,2,3,4,5,6,7$, which gives you seven quadratic equations to check


• for example, $x=0$ is a solution when $n=1$

share|cite|improve this answer

answered Jul 18 at 9:55

Henry

93.1k469147

add a comment | 

up vote
1
down vote

Well, first of all you're checking only 'all' integers and not all reals which include also $1/2$ or $pi$ for example. I will try to give you an outline of how I would approach it.

Second of all, note that the left-hand side is always between $0$ and $1$. So, what is implied about the range $x$ might be in?

Once, you determined that you should be splitting up the range of possible $x$ into smaller segments such that for each segment you can find a natural number $n$ such that
$$(x+1)^3 = (x+1)^3 - n.$$
Hint: there are $7$ segments that you should consider. Then for each segment it boils down in solving
$$ (x+1)^3 - n = x^3.$$
This looks difficult at first sight but I guarantee it will simplify to a quadratic polynomial (which you can solve). Please don't forget that the solutions of this equation should still be in the range on which we consider solutions to be part of.

I hope this helps and if you have any questions about any of these steps, feel free to ask them.

share|cite|improve this answer

edited Jul 18 at 12:36

answered Jul 18 at 10:02

Stan Tendijck

1,277110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the explanation ... This really helped. ...
  – Math Tise
  Jul 18 at 10:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I get the following answers :
  – Math Tise
  Jul 18 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=0, x is imaginary, for n=1 to 6, i get $x$ = $0$, $fracsqrt21-3 6$, $fracsqrt33-3 6$, $fracsqrt5-1 2$, $fracsqrt57-3 6$, $fracsqrt69-3 6$ .
  – Math Tise
  Jul 18 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=7, x is a whole number, for n=8 to 17, i get $fracsqrt93-3 6$, $fracsqrt105-3 6$, $fracsqrt13-1 2$, $fracsqrt129-3 6$, $fracsqrt141-3 6$, $fracsqrt17-1 2$ .
  – Math Tise
  Jul 18 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the list goes on ... are there only particular solutions to this problem?, or unlimited solutions... please advise @Stan Tendijck ... Thanks again.
  – Math Tise
  Jul 18 at 10:40
  

  
  
  
  

 | 
show 2 more comments

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

Fractional part is always in $[0,1)$. So the domain of $g(x)=x^3$ you are looking for is such that $R_gin[0,1)$, which happens to be $[0,1)$.

$y=(x+1)^3$ is basically $y=(x+1)^3$ chopped into appropriate pieces and translated down by some $kinmathbbZ$ where $k=lfloor(x+1)^3rfloor$

So we solve
$$(x+1)^3-k=x^3$$
which gives
$$k=3x^2+3x+1$$

Since for $[0,1)$, $0le(x+1)^3<8$, we check
$$3x^2+3x+1=0,1,2,3,4,5,6,7$$
whose positive roots give the answers.

share|cite|improve this answer

edited Jul 18 at 10:06

answered Jul 18 at 9:55

Karn Watcharasupat

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact the fractional part is always in $[0,1)$
  – MPW
  Jul 18 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MPW ugh my bad! edited. thanks a lot
  – Karn Watcharasupat
  Jul 18 at 9:57
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

Fractional part is always in $[0,1)$. So the domain of $g(x)=x^3$ you are looking for is such that $R_gin[0,1)$, which happens to be $[0,1)$.

$y=(x+1)^3$ is basically $y=(x+1)^3$ chopped into appropriate pieces and translated down by some $kinmathbbZ$ where $k=lfloor(x+1)^3rfloor$

So we solve
$$(x+1)^3-k=x^3$$
which gives
$$k=3x^2+3x+1$$

Since for $[0,1)$, $0le(x+1)^3<8$, we check
$$3x^2+3x+1=0,1,2,3,4,5,6,7$$
whose positive roots give the answers.

share|cite|improve this answer

edited Jul 18 at 10:06

answered Jul 18 at 9:55

Karn Watcharasupat

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact the fractional part is always in $[0,1)$
  – MPW
  Jul 18 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MPW ugh my bad! edited. thanks a lot
  – Karn Watcharasupat
  Jul 18 at 9:57
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

Fractional part is always in $[0,1)$. So the domain of $g(x)=x^3$ you are looking for is such that $R_gin[0,1)$, which happens to be $[0,1)$.

$y=(x+1)^3$ is basically $y=(x+1)^3$ chopped into appropriate pieces and translated down by some $kinmathbbZ$ where $k=lfloor(x+1)^3rfloor$

So we solve
$$(x+1)^3-k=x^3$$
which gives
$$k=3x^2+3x+1$$

Since for $[0,1)$, $0le(x+1)^3<8$, we check
$$3x^2+3x+1=0,1,2,3,4,5,6,7$$
whose positive roots give the answers.

share|cite|improve this answer

edited Jul 18 at 10:06

answered Jul 18 at 9:55

Karn Watcharasupat

3,8192426

Fractional part is always in $[0,1)$. So the domain of $g(x)=x^3$ you are looking for is such that $R_gin[0,1)$, which happens to be $[0,1)$.

$y=(x+1)^3$ is basically $y=(x+1)^3$ chopped into appropriate pieces and translated down by some $kinmathbbZ$ where $k=lfloor(x+1)^3rfloor$

So we solve
$$(x+1)^3-k=x^3$$
which gives
$$k=3x^2+3x+1$$

Since for $[0,1)$, $0le(x+1)^3<8$, we check
$$3x^2+3x+1=0,1,2,3,4,5,6,7$$
whose positive roots give the answers.

share|cite|improve this answer

edited Jul 18 at 10:06

answered Jul 18 at 9:55

Karn Watcharasupat

3,8192426

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 10:06

edited Jul 18 at 10:06

edited Jul 18 at 10:06

answered Jul 18 at 9:55

Karn Watcharasupat

3,8192426

answered Jul 18 at 9:55

Karn Watcharasupat

3,8192426

answered Jul 18 at 9:55

Karn Watcharasupat

3,8192426

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact the fractional part is always in $[0,1)$
  – MPW
  Jul 18 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MPW ugh my bad! edited. thanks a lot
  – Karn Watcharasupat
  Jul 18 at 9:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact the fractional part is always in $[0,1)$
  – MPW
  Jul 18 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MPW ugh my bad! edited. thanks a lot
  – Karn Watcharasupat
  Jul 18 at 9:57
  

  
  
  
  

In fact the fractional part is always in $[0,1)$
– MPW
Jul 18 at 9:55

In fact the fractional part is always in $[0,1)$
– MPW
Jul 18 at 9:55

@MPW ugh my bad! edited. thanks a lot
– Karn Watcharasupat
Jul 18 at 9:57

@MPW ugh my bad! edited. thanks a lot
– Karn Watcharasupat
Jul 18 at 9:57

add a comment | 

up vote
5
down vote

Hints:

• $0 le y lt 1$


• so any solution to $(x +1)^3 = x^3$ has $0 le x^3 lt 1$ and thus $0 le x lt 1$


• so $1 le x+1 lt 2$ and $1 le (x+1)^3 lt 8$


• any solution has $(x+1)^3 = x^3 +n$ for $n in 1,2,3,4,5,6,7$, which gives you seven quadratic equations to check


• for example, $x=0$ is a solution when $n=1$

share|cite|improve this answer

answered Jul 18 at 9:55

Henry

93.1k469147

add a comment | 

up vote
5
down vote

Hints:

• $0 le y lt 1$


• so any solution to $(x +1)^3 = x^3$ has $0 le x^3 lt 1$ and thus $0 le x lt 1$


• so $1 le x+1 lt 2$ and $1 le (x+1)^3 lt 8$


• any solution has $(x+1)^3 = x^3 +n$ for $n in 1,2,3,4,5,6,7$, which gives you seven quadratic equations to check


• for example, $x=0$ is a solution when $n=1$

share|cite|improve this answer

answered Jul 18 at 9:55

Henry

93.1k469147

add a comment | 

up vote
5
down vote

up vote
5
down vote

Hints:

• $0 le y lt 1$


• so any solution to $(x +1)^3 = x^3$ has $0 le x^3 lt 1$ and thus $0 le x lt 1$


• so $1 le x+1 lt 2$ and $1 le (x+1)^3 lt 8$


• any solution has $(x+1)^3 = x^3 +n$ for $n in 1,2,3,4,5,6,7$, which gives you seven quadratic equations to check


• for example, $x=0$ is a solution when $n=1$

share|cite|improve this answer

answered Jul 18 at 9:55

Henry

93.1k469147

Hints:

• $0 le y lt 1$


• so any solution to $(x +1)^3 = x^3$ has $0 le x^3 lt 1$ and thus $0 le x lt 1$


• so $1 le x+1 lt 2$ and $1 le (x+1)^3 lt 8$


• any solution has $(x+1)^3 = x^3 +n$ for $n in 1,2,3,4,5,6,7$, which gives you seven quadratic equations to check


• for example, $x=0$ is a solution when $n=1$

share|cite|improve this answer

answered Jul 18 at 9:55

Henry

93.1k469147

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 9:55

Henry

93.1k469147

answered Jul 18 at 9:55

Henry

93.1k469147

answered Jul 18 at 9:55

Henry

93.1k469147

93.1k469147

add a comment | 

add a comment | 

up vote
1
down vote

Well, first of all you're checking only 'all' integers and not all reals which include also $1/2$ or $pi$ for example. I will try to give you an outline of how I would approach it.

Second of all, note that the left-hand side is always between $0$ and $1$. So, what is implied about the range $x$ might be in?

Once, you determined that you should be splitting up the range of possible $x$ into smaller segments such that for each segment you can find a natural number $n$ such that
$$(x+1)^3 = (x+1)^3 - n.$$
Hint: there are $7$ segments that you should consider. Then for each segment it boils down in solving
$$ (x+1)^3 - n = x^3.$$
This looks difficult at first sight but I guarantee it will simplify to a quadratic polynomial (which you can solve). Please don't forget that the solutions of this equation should still be in the range on which we consider solutions to be part of.

I hope this helps and if you have any questions about any of these steps, feel free to ask them.

share|cite|improve this answer

edited Jul 18 at 12:36

answered Jul 18 at 10:02

Stan Tendijck

1,277110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the explanation ... This really helped. ...
  – Math Tise
  Jul 18 at 10:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I get the following answers :
  – Math Tise
  Jul 18 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=0, x is imaginary, for n=1 to 6, i get $x$ = $0$, $fracsqrt21-3 6$, $fracsqrt33-3 6$, $fracsqrt5-1 2$, $fracsqrt57-3 6$, $fracsqrt69-3 6$ .
  – Math Tise
  Jul 18 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=7, x is a whole number, for n=8 to 17, i get $fracsqrt93-3 6$, $fracsqrt105-3 6$, $fracsqrt13-1 2$, $fracsqrt129-3 6$, $fracsqrt141-3 6$, $fracsqrt17-1 2$ .
  – Math Tise
  Jul 18 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the list goes on ... are there only particular solutions to this problem?, or unlimited solutions... please advise @Stan Tendijck ... Thanks again.
  – Math Tise
  Jul 18 at 10:40
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

Well, first of all you're checking only 'all' integers and not all reals which include also $1/2$ or $pi$ for example. I will try to give you an outline of how I would approach it.

Second of all, note that the left-hand side is always between $0$ and $1$. So, what is implied about the range $x$ might be in?

Once, you determined that you should be splitting up the range of possible $x$ into smaller segments such that for each segment you can find a natural number $n$ such that
$$(x+1)^3 = (x+1)^3 - n.$$
Hint: there are $7$ segments that you should consider. Then for each segment it boils down in solving
$$ (x+1)^3 - n = x^3.$$
This looks difficult at first sight but I guarantee it will simplify to a quadratic polynomial (which you can solve). Please don't forget that the solutions of this equation should still be in the range on which we consider solutions to be part of.

I hope this helps and if you have any questions about any of these steps, feel free to ask them.

share|cite|improve this answer

edited Jul 18 at 12:36

answered Jul 18 at 10:02

Stan Tendijck

1,277110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the explanation ... This really helped. ...
  – Math Tise
  Jul 18 at 10:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I get the following answers :
  – Math Tise
  Jul 18 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=0, x is imaginary, for n=1 to 6, i get $x$ = $0$, $fracsqrt21-3 6$, $fracsqrt33-3 6$, $fracsqrt5-1 2$, $fracsqrt57-3 6$, $fracsqrt69-3 6$ .
  – Math Tise
  Jul 18 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=7, x is a whole number, for n=8 to 17, i get $fracsqrt93-3 6$, $fracsqrt105-3 6$, $fracsqrt13-1 2$, $fracsqrt129-3 6$, $fracsqrt141-3 6$, $fracsqrt17-1 2$ .
  – Math Tise
  Jul 18 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the list goes on ... are there only particular solutions to this problem?, or unlimited solutions... please advise @Stan Tendijck ... Thanks again.
  – Math Tise
  Jul 18 at 10:40
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

up vote
1
down vote

Well, first of all you're checking only 'all' integers and not all reals which include also $1/2$ or $pi$ for example. I will try to give you an outline of how I would approach it.

Second of all, note that the left-hand side is always between $0$ and $1$. So, what is implied about the range $x$ might be in?

Once, you determined that you should be splitting up the range of possible $x$ into smaller segments such that for each segment you can find a natural number $n$ such that
$$(x+1)^3 = (x+1)^3 - n.$$
Hint: there are $7$ segments that you should consider. Then for each segment it boils down in solving
$$ (x+1)^3 - n = x^3.$$
This looks difficult at first sight but I guarantee it will simplify to a quadratic polynomial (which you can solve). Please don't forget that the solutions of this equation should still be in the range on which we consider solutions to be part of.

I hope this helps and if you have any questions about any of these steps, feel free to ask them.

share|cite|improve this answer

edited Jul 18 at 12:36

answered Jul 18 at 10:02

Stan Tendijck

1,277110

Well, first of all you're checking only 'all' integers and not all reals which include also $1/2$ or $pi$ for example. I will try to give you an outline of how I would approach it.

Second of all, note that the left-hand side is always between $0$ and $1$. So, what is implied about the range $x$ might be in?

Once, you determined that you should be splitting up the range of possible $x$ into smaller segments such that for each segment you can find a natural number $n$ such that
$$(x+1)^3 = (x+1)^3 - n.$$
Hint: there are $7$ segments that you should consider. Then for each segment it boils down in solving
$$ (x+1)^3 - n = x^3.$$
This looks difficult at first sight but I guarantee it will simplify to a quadratic polynomial (which you can solve). Please don't forget that the solutions of this equation should still be in the range on which we consider solutions to be part of.

I hope this helps and if you have any questions about any of these steps, feel free to ask them.

share|cite|improve this answer

edited Jul 18 at 12:36

answered Jul 18 at 10:02

Stan Tendijck

1,277110

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 12:36

edited Jul 18 at 12:36

edited Jul 18 at 12:36

answered Jul 18 at 10:02

Stan Tendijck

1,277110

answered Jul 18 at 10:02

Stan Tendijck

1,277110

answered Jul 18 at 10:02

Stan Tendijck

1,277110

1,277110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the explanation ... This really helped. ...
  – Math Tise
  Jul 18 at 10:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I get the following answers :
  – Math Tise
  Jul 18 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=0, x is imaginary, for n=1 to 6, i get $x$ = $0$, $fracsqrt21-3 6$, $fracsqrt33-3 6$, $fracsqrt5-1 2$, $fracsqrt57-3 6$, $fracsqrt69-3 6$ .
  – Math Tise
  Jul 18 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=7, x is a whole number, for n=8 to 17, i get $fracsqrt93-3 6$, $fracsqrt105-3 6$, $fracsqrt13-1 2$, $fracsqrt129-3 6$, $fracsqrt141-3 6$, $fracsqrt17-1 2$ .
  – Math Tise
  Jul 18 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the list goes on ... are there only particular solutions to this problem?, or unlimited solutions... please advise @Stan Tendijck ... Thanks again.
  – Math Tise
  Jul 18 at 10:40
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the explanation ... This really helped. ...
  – Math Tise
  Jul 18 at 10:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I get the following answers :
  – Math Tise
  Jul 18 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=0, x is imaginary, for n=1 to 6, i get $x$ = $0$, $fracsqrt21-3 6$, $fracsqrt33-3 6$, $fracsqrt5-1 2$, $fracsqrt57-3 6$, $fracsqrt69-3 6$ .
  – Math Tise
  Jul 18 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for n=7, x is a whole number, for n=8 to 17, i get $fracsqrt93-3 6$, $fracsqrt105-3 6$, $fracsqrt13-1 2$, $fracsqrt129-3 6$, $fracsqrt141-3 6$, $fracsqrt17-1 2$ .
  – Math Tise
  Jul 18 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the list goes on ... are there only particular solutions to this problem?, or unlimited solutions... please advise @Stan Tendijck ... Thanks again.
  – Math Tise
  Jul 18 at 10:40
  

  
  
  
  

Thanks for the explanation ... This really helped. ...
– Math Tise
Jul 18 at 10:26

Thanks for the explanation ... This really helped. ...
– Math Tise
Jul 18 at 10:26

I get the following answers :
– Math Tise
Jul 18 at 10:27

I get the following answers :
– Math Tise
Jul 18 at 10:27

for n=0, x is imaginary, for n=1 to 6, i get $x$ = $0$, $fracsqrt21-3 6$, $fracsqrt33-3 6$, $fracsqrt5-1 2$, $fracsqrt57-3 6$, $fracsqrt69-3 6$ .
– Math Tise
Jul 18 at 10:37

for n=0, x is imaginary, for n=1 to 6, i get $x$ = $0$, $fracsqrt21-3 6$, $fracsqrt33-3 6$, $fracsqrt5-1 2$, $fracsqrt57-3 6$, $fracsqrt69-3 6$ .
– Math Tise
Jul 18 at 10:37

for n=7, x is a whole number, for n=8 to 17, i get $fracsqrt93-3 6$, $fracsqrt105-3 6$, $fracsqrt13-1 2$, $fracsqrt129-3 6$, $fracsqrt141-3 6$, $fracsqrt17-1 2$ .
– Math Tise
Jul 18 at 10:39

for n=7, x is a whole number, for n=8 to 17, i get $fracsqrt93-3 6$, $fracsqrt105-3 6$, $fracsqrt13-1 2$, $fracsqrt129-3 6$, $fracsqrt141-3 6$, $fracsqrt17-1 2$ .
– Math Tise
Jul 18 at 10:39

and the list goes on ... are there only particular solutions to this problem?, or unlimited solutions... please advise @Stan Tendijck ... Thanks again.
– Math Tise
Jul 18 at 10:40

and the list goes on ... are there only particular solutions to this problem?, or unlimited solutions... please advise @Stan Tendijck ... Thanks again.
– Math Tise
Jul 18 at 10:40

 | 
show 2 more comments

 

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