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No. Here is a proof by counterexample that it does not.

Given ab + bc + ca = 3:

Assume toward a contradiction that abc is a cube. Then a = b = c.

Without loss of generality, let a = 2, b = 2, and c = 2.

Then ab = 4, bc = 4, and ca = 4.

ab + bc + ca = 4 + 4 + 4 = 12.

Therefore, 12 = 3, which is false, and so the original statement is false.

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