SOLUTION: Hopefully someone can point out where i am going wrong: {{{x^6-9x^3+8=0}}} -- I let U = x^3 -- {{{U^2 -9U +8 = 0}}} -- {{{U^2-9U+(81/4) = (49/4)}}} -- {{{(2U-9)^2/4 = 49

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Hopefully someone can point out where i am going wrong: {{{x^6-9x^3+8=0}}} -- I let U = x^3 -- {{{U^2 -9U +8 = 0}}} -- {{{U^2-9U+(81/4) = (49/4)}}} -- {{{(2U-9)^2/4 = 49      Log On


   

Question 560674: Hopefully someone can point out where i am going wrong:
x%5E6-9x%5E3%2B8=0
--
I let U = x^3
--
U%5E2+-9U+%2B8+=+0
--
U%5E2-9U%2B%2881%2F4%29+=+%2849%2F4%29
--
%282U-9%29%5E2%2F4+=+49%2F4
--
2U-9 = +/- 7
--
U = 1, 8
---
Sub x^3 back for U
---
x%5E3+=+1 x=1
---
x%5E3+=+8 x=2
---
Problem is the book shows the answer to be:
-3%2F2 +/- i+%2A+sqrt%283%29%2F2

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The mistake occurred when you had x^3 = 1 and x^3 = 8 and concluded that the roots were 1 and 2. However x^3 = 1 has three roots of unity in the complex plane, namely







The roots for x^3 = 8 are similar, except twice as much (e.g. each root of x^3 = 1, multiplied by 2).