SOLUTION: I have to find all roots of X^6 + 9X^3 + 8 = 0. I first let Y=X^3, then my equation became Y^2 + 9Y + 8 = 0. Solving for Y, I have Y = -8 OR -1. So I have two possible roots as X

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I have to find all roots of X^6 + 9X^3 + 8 = 0. I first let Y=X^3, then my equation became Y^2 + 9Y + 8 = 0. Solving for Y, I have Y = -8 OR -1. So I have two possible roots as X      Log On


   

Question 231979: I have to find all roots of X^6 + 9X^3 + 8 = 0.
I first let Y=X^3, then my equation became Y^2 + 9Y + 8 = 0. Solving for Y, I have Y = -8 OR -1.
So I have two possible roots as X = -2 and X = -1.
Since the highest power in the equation is 6, there are four other roots.
How do you find the other roots?
Thanks,
Alan
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E6+%2B+9x%5E3+%2B+8+=+0
I took a direct approach and factored it into
%28x%5E3+%2B+8%29%28x%5E3+%2B+1%29+=+0
x%5E3+=+-8 is a solution
x+=+-2
I need the other 2 roots now. I proceed to cheat and
apply what I know about complex roots, namely that
they are equally spaced aound the complex plane. That means
the 3 roots are 120 degrees apart. One complex root is
30 degrees from the vertical in 1st quadrant, the other is
30 degrees from vertical in 4th quadrant. The components of
the complex roots are
x+=+1+%2B+sqrt%283%29i (1st quadrant) and
x+=+1+-+sqrt%283%29i (4th quadrant).
I do the same for x%5E3+%2B+1
x%5E3+=+-1
The complex roots are
x+=+1%2F2+%2B+%28sqrt%283%29%2F2%29%2Ai (1st quadrant) and
x+=+1%2F2+-+%28sqrt%283%29%2F2%29%2Ai (4th quadrant
I don't know an easy way to prove that's right, but
I'm pretty sure it is