Question

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

Answer 1

Answer: x=1.6988

Step-by-step explanation:

Answer 2

Answer:

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