How do you find all the real and complex roots of #x^3 + 8 = 0#?
1 Answer
Jan 19, 2016
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
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#
