Question

Find the zeros of the function g(x)=x^3-3x^2+3x-1

Answer 1

if you look closely,

`{x}^(3) - 3 {x}^(2) + 3x - 1`

is actually a perfect cube

`(a-b)^3 = a^3-3a^2b+3ab^2-b^3`

`{(x)}^(3) + 3 {(x)}^(2) ( - 1) + 3(x) {( - 1)}^(2) + {( - 1)}^(3)`

so,

(x-1)^3

hence, all three roots are equal which is 1

Answer 2

Answer: x=1.6988

Explanation: