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y to the six is common, so it becomes y^3(y^3 - 125). Then we use difference of cubes ((a-b)(a^2 - 2ab + b^2)), so it becomes y^3(y-5)(y^2 + 5y + 25). Since we cannot factor out that last quadratic part, this concludes the factoring process and we end up with that last line as our answer.

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13y ago
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Q: How do you factor y to the sixth power minus 125y cubed?
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