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:

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

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