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This question already has an answer here:

Reduce[x^2 + y^3 == z && x + 2 y == 3 z + 1 && x y z != 0, {x, y, z}]

you get (y == Root[1 - x + 3 x^2 - 2 #1 + 3 #1^3 &, 1] || y == Root[1 - x + 3 x^2 - 2 #1 + 3 #1^3 &, 2] || y == Root[1 - x + 3 x^2 - 2 #1 + 3 #1^3 &, 3]) && z == 1/3 (-1 + x + 2 y) && -x y + x^2 y + 2 x y^2 != 0

How can we get rid of #?

Just like Reduce[6 - 3 x^2 == 2 n^2 + 4 n x, x]

x == 1/3 (-2 n - Sqrt[2] Sqrt[9 - n^2]) || x == 1/3 (-2 n + Sqrt[2] Sqrt[9 - n^2])

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marked as duplicate by Kuba May 29 at 5:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Wrap the output with ToRadicals:

ToRadicals @ Reduce[x^2 + y^3 == z && x + 2 y == 3 z + 1 && x y z != 0, {x, y, z}]

enter image description here

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