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Unformatted text preview: .v 3 ± u 3 / C q D 3uv D p: Now, using the second equation to eliminate u from the ﬁrst gives a quadratic equation for v 3 : v 3 ± p 3 27v 3 C q D 0: The solutions to this are v 3 D ± q 2 ˙ s q 2 4 C p 3 27 : It turns out that we get the same solutions to our original equation regardless of the choice of the square root. Let ! denote the primitive third root of unity ! D e 2±i=3 , and let A denote one cube root of ± q 2 C s q 2 4 C p 3 27 : Then the solutions for v are A; !A , and ! 2 A , and the solutions for x D v ± p 3v are ˛ 1 D A ± p 3A ; a 2 D !A ± ! 2 p 3A ; ˛ 3 D ! 2 A ± ! p 3A :...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
 Fall '08
 EVERAGE
 Algebra

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