Questions on Algebra: Square root, cubic root, N-th root answered by real tutors!

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Question 151868: how many roots does x^6-28x^3+27=0 have? (hint: trinomial expression, use y = x^3): how many roots does x^6-28x^3+27=0 have? (hint: trinomial expression, use y = x^3)
Answer by Earlsdon(3719) About Me  (Show Source):
You can put this solution on YOUR website!
x^6-28x^3+27 = 0 Substitute y = x^3
y^2-28y+27 = 0 Factor the trinomial.
(y-27)(y-1) = 0
Roots for this are:
y = 27 and y = 1 Substitute x^3 = y
x^3 = 27 and x^3 = 1 Take the cube root.
x = 3 and x = 1
The roots are: x = 1 and x = 3
graph(400,400,-5,5,-200,300,x^6-28x^3+27)