SOLUTION: Find all the roots of : a) (x^3) - 1=0 b)(x^5)- 32j=0

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Question 177514: Find all the roots of :
a) (x^3) - 1=0
b)(x^5)- 32j=0

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find all the roots of :
a) (x^3) - 1=0
Factor to get:
(x-1)(x^2 + x + 1) = 0
x = 1, or x = [-1 +- sqrt(1 - 4*1*1)]/2 = (-1+sqrt(3)i)/2 or (-1-sqrt(3)i)/2
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b)(x^5)- 32j=0
x^5 - (2i)^ = 0
(x-2i)(?) = 0
? is a 4th power polynomial with complex coefficients.
Zero: x = 2i and 4 others
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Cheers,
Stan H.