SOLUTION: how many roots does x^6-28x^3+27=0 have? (hint: trinomial expression, use y = x^3)

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Question 151868: how many roots does x^6-28x^3+27=0 have? (hint: trinomial expression, use y = x^3)
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E6-28x%5E3%2B27+=+0 Substitute y+=+x%5E3
y%5E2-28y%2B27+=+0 Factor the trinomial.
%28y-27%29%28y-1%29+=+0
Roots for this are:
y+=+27 and y+=+1 Substitute x%5E3+=+y
x%5E3+=+27 and x%5E3+=+1 Take the cube root.
x+=+3 and x+=+1
The roots are: x+=+1 and x+=+3
graph%28400%2C400%2C-5%2C5%2C-200%2C300%2Cx%5E6-28x%5E3%2B27%29