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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.

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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.

May 13, 2015

#1
+98173
+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

May 13, 2015

#1
+98173
+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
#2
+99352
0

Great job there Chris

May 13, 2015
#3
+98173
0

Thanks, Melody.....

May 13, 2015