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8 July, 10:28

What are the real and complex solutions of the polynomial equation x^3-64=0

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  1. 8 July, 10:54
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    I don't understand at all. I know to use the "difference" of cubes" formula, a3 + b3 = (a+b) (a2 - ab + b2) to simplify. Which I did. And then simplify. And then that makes (x+64) a factor and - 64 a root (I think?). So far, I have: (x) ^3+64^ (3) (x+64) (x^2-64x+64^2) (x+64) (x^2-64x+4096) x=-64 (?) so it would be - 64 i would say.
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