How do you find all the real and complex roots of #x^3 + 8 = 0#?

1 Answer
Jan 19, 2016

Answer:

#x=-2,1+-sqrt3i#

Explanation:

Split this apart using the sum of cubes identity.

#(x+2)(x^2-2x+4)=0#

Now, you know that either

#x+2=0color(white)(xxxx)"or"color(white)(xxxx)x^2-2x+4=0#

Solving the first of these gives that #x=-2#.

The other two can be found through applying the quadratic formula on the quadratic:

#x=(2+-sqrt(4-16))/2=(2+-2sqrt3i)/2=1+-sqrt3i#