How do you find all the real and complex roots of # x^6-28x^3+27=0#?

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Answer 1 · 2016-01-22T22:06:31

Notice that the sum of the coefficients is zero to factor as #(x^3-1)(x^3-27)#, then identify all of the cube roots.

The primitive Complex cube root of #1# is:

#omega = -1/2 +sqrt(3)/2i#

The cube roots of #1# are #1#, #omega# and #omega^2 = bar(omega)#

To verify this you can factor #x^3-1 = (x-1)(x^2+x+1)# then use the quadratic formula to find that the zeros of #x^2+x+1# are:

#-1/2+-sqrt(3)/2i#

#x^6-28x^3+27# is a quadratic in #x^3# with coefficients whose sum is #0#. Hence #(x^3-1)# is a factor and we find:

#x^6-28x^3+27#

#=(x^3-1)(x^3-27)#

#=(x^3-1^3)(x^3-3^3)#

#=(x-1)(x-omega)(x-omega^2)(x-3)(x-3omega)(x-3omega^2)#

So the roots of #x^6-28x^3+27 = 0# are:

#x=1#

#x=omega = -1/2+sqrt(3)/2i#

#x=omega^2 = -1/2-sqrt(3)/2i#

#x=3#

#x=3omega = -3/2+(3sqrt(3))/2i#

#x=3omega^2 = -3/2-(3sqrt(3))/2i#