How do you find all the real and complex roots of #f(x)= x^3-27#?

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Answer 1 · 2016-02-06T02:54:53

#x=3,(-3+-3isqrt3)/2#

The roots occur when the function equals #0#.

#x^3-27=0#

Note that #x^3-27# can be factored as a difference of cubes.

#(x-3)(x^2+3x+9)=0#

The #(x-3)# term yields a root of #x=3#.

The roots of #(x^3+3x+9)# can be found through the quadratic equation:

#x=(-3+-sqrt(9-36))/2=(-3+-sqrt(-27))/2=(-3+-3isqrt3)/2#