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2015-04-21T01:28:37+05:30

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Let x^3 = y

   y^2  + 5 y + 8
  let us find roots of    y^2 + 5 y + 8 = 0
  discriminant :     squareroot(25 - 40) ]
  as the discriminant is negative, there are no real roots. 
  The given expression canno t be  factorized with real  factors.

     y1 = [- 5 + √15 i ] / 2              y2 = [ -5 - √15 ] / 2
 are the solutions which involve imaginary numbers.

      x⁶ + 5 x³ + 8  =     (2 x³ + 5 - √15 i) (2x³ + 5 - √15 i) / 4

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2015-04-21T05:54:19+05:30

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This polynomial is specially made for using the identity
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

With the idea of using above identity, split the middle term 5*x^3 as (-x)^3+6x^3=(-x)^3-3(x^2)(-x)(2) :

x^6+5x^3+8=(x^2)^3+(-x)^3+2^3-3(x^2)(-x)(2)\\=(x^2-x+2)(x^4+x^2+4+x^3+2x-2x^2)\\=(x^2-x+2)(x^4+x^3-x^2+2x+4)
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