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

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