SOLUTION: Hi I am having trouble solving the following problem: x^3-8=0 Can you provide some help in solving this - there are three answers one of which is 2 but I cannot figure out the

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Hi I am having trouble solving the following problem: x^3-8=0 Can you provide some help in solving this - there are three answers one of which is 2 but I cannot figure out the      Log On


   

Question 8979: Hi
I am having trouble solving the following problem:
x^3-8=0
Can you provide some help in solving this - there are three answers one of which is 2 but I cannot figure out the other two imaginary ones.
Thanks
Lee
Found 2 solutions by longjonsilver, rapaljer:
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
there should be an identity to factorise x%5E3+-+a%5E3 ... you have x%5E3-2%5E3.

I just cannot lay my hands on it at the moment...that will give you (x-2) and a quadratic bracket...solve that to give you your 2 imaginary answers

jon.

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E3+-+8+=+0 Factors into
+%28x-2%29%28x%5E2+%2B+2x+%2B+4%29+=+0+

The quadratic factor can be solved by either the quadratic formula, the pluggable solver, or by completing the square. Completing the square is fairly easy:
x%5E2+%2B+2x+%2B+_____+=+-4+%2B+____+
x%5E2+%2B+2x+%2B+1+++++=+-4+%2B+1+
+%28x%2B1%29%5E2+=+-3+
++%28x%2B1%29+=+sqrt%28-3%29+
+x+=+-1+%2B-+i+%2Asqrt%283%29+, and of course, x=2!!

R^2