perplexus.info :: Just Math : From AP to cubes

From AP to cubes (Posted on 2024-12-07)

Let a, b, and c be positive integers such that a^1/3, b^1/3, and c^1/3 are terms of an arithmetic sequence. Prove that abc is a perfect cube.

No Solution Yet Submitted by Danish Ahmed Khan

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Solution?

| Comment 1 of 3

The premise of the problem may not be correct

a^1/3=b^1/3-d

c^1/3=b^1/3+d

(abc)^1/3=(b^1/3-d)b(b^1/3+d)

abc = (b^1/3-d)³b(b^1/3+d)³

=(b-b^1/3d²)³

Which works as long as b is a perfect cube.

Here's an example showing it doesn't work if b is not a cube.

Suppose b=10, d=1

a^1/3=1.1544

b^1/3=2.1544

c^1/3=3.1544

a^1/3b^1/3c^1/3=7.8455

abc=482.9172

Which is not a perfect cube but does equal (b-b^1/3d²)³

Posted by Jer on 2024-12-07 13:37:03

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