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Find all sets of four positive consecutive integers such that the sum of the cubes of the first three is the cube of the fourth.

Guest May 13, 2015

Best Answer 

 #1
avatar+78750 
+10

We have

x^3 + (x + 1)^3 + (x + 2)^3  = (x + 3)^3  ...  and expanding, we have

x^3 + x^3 + 3 x^2 + 3x + 1 + x^3 + 6x^2 + 12x + 8  = x^3 + 9x^2 + 27x + 27  ....  simplify

3x^3 + 9x^2 + 15x + 9  = x^3 + 9x^2 + 27x + 27

2x^3 - 12x - 18 = 0

x^3 - 6x - 9 = 0   Using the Rational Zeroes Theorem, the possible roots of this are ±1,  ±3  and  ±9 

And  x = 3 is one solution

Using synthetic division. we have

3        1    0   - 6    -9

                3     9     9

          1    3    3      0

 

And the remaining polynomial  is   x^2 + 3x + 3   which has no "real" roots

 

So the integers are   3,4, 5    and 6

 

Check.....    3^3 + 4^3 + 5^3   =  27 + 64 + 125 = 216  = 6^3

 

 

  

CPhill  May 13, 2015
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3+0 Answers

 #1
avatar+78750 
+10
Best Answer

We have

x^3 + (x + 1)^3 + (x + 2)^3  = (x + 3)^3  ...  and expanding, we have

x^3 + x^3 + 3 x^2 + 3x + 1 + x^3 + 6x^2 + 12x + 8  = x^3 + 9x^2 + 27x + 27  ....  simplify

3x^3 + 9x^2 + 15x + 9  = x^3 + 9x^2 + 27x + 27

2x^3 - 12x - 18 = 0

x^3 - 6x - 9 = 0   Using the Rational Zeroes Theorem, the possible roots of this are ±1,  ±3  and  ±9 

And  x = 3 is one solution

Using synthetic division. we have

3        1    0   - 6    -9

                3     9     9

          1    3    3      0

 

And the remaining polynomial  is   x^2 + 3x + 3   which has no "real" roots

 

So the integers are   3,4, 5    and 6

 

Check.....    3^3 + 4^3 + 5^3   =  27 + 64 + 125 = 216  = 6^3

 

 

  

CPhill  May 13, 2015
 #2
avatar+91049 
0

Great job there Chris   

Melody  May 13, 2015
 #3
avatar+78750 
0

Thanks, Melody.....

 

  

CPhill  May 13, 2015

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