Answer
$x=2$
Work Step by Step
$x^3-6x^2+12x-8=0$
$x^3-4x^2+4x-2x^2+8x-8=x(x^2-4x+4)-2(x^2-4x+4)=(x-2)(x^2-4x+4)$
$(x-2)(x^2-4x+4)=(x-2)(x^2-2x-2x+4)=(x-2)(x(x-2)-2(x-2))=(x-2)(x-2)(x-2)$
The possible real solutions of the equation of
$(x-2)^3=0$ is:
$x=2$
