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Find the positive integer whose cube exceeds its square by 4624.

 Dec 8, 2019
 #1
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The number is:

 

17^3  -  17^2 =4,624

 Dec 8, 2019
 #2
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+1

Solve for a:
a^3 - a^2 = 4624

Subtract 4624 from both sides:
a^3 - a^2 - 4624 = 0

The left hand side factors into a product with two terms:
(a - 17) (a^2 + 16 a + 272) = 0

Split into two equations:
a - 17 = 0 or a^2 + 16 a + 272 = 0

Add 17 to both sides:
a = 17 or a^2 + 16 a + 272 = 0

Subtract 272 from both sides:
a = 17 or a^2 + 16 a = -272

Add 64 to both sides:
a = 17 or a^2 + 16 a + 64 = -208

Write the left hand side as a square:
a = 17 or (a + 8)^2 = -208

Take the square root of both sides:
a = 17 or a + 8 = 4 i sqrt(13) or a + 8 = -4 i sqrt(13)

Subtract 8 from both sides:
a = 17 or a = 4 i sqrt(13) - 8 or a + 8 = -4 i sqrt(13)

Subtract 8 from both sides:


a = 17 

 Dec 8, 2019

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