How do you find the inverse of #p(x)=x^3-3x^2+3x-1#?

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Answer 1 · 2015-12-28T10:11:15

Find #p(x) = (x-1)^3#, hence:

#p^(-1)(y) = root(3)(y)+1#

Let #y = p(x) = x^3-3x^2+3x-1 = (x-1)^3#

Taking the cube root of both ends we find:

#root(3)(y) = x-1#

Add #1# to both sides and transpose to get:

#x = root(3)(y)+1#

Hence #p^(-1)(y) = root(3)(y)+1#