Given: f(x)=x^5-12x^4+54x^3-108x^2+81x=0 ···Eq.1
Let f(x)=xg(x)=0, so x=0 or g(x)=0 and g(x)=x^4-12x^3+54x^2-108x+81 Here, we examine g(x)=0.
Notice that the coefficients of each term in g(x) are 1,-12,54,-108,81, so the greatest common factor of 54,-108, 81 is 27, the sum of 2 factors of 27, 3 and 9, is 12, and the product of 3 and 27 is 81. Thus, We divide g(x) into 2 parts assuming (x-3) might be one of the factors, getting
g(x)=(x^4-12x^3+27x^2)+(27x^2-108x+81)=x^2(x^2-12x+27)+27(x^2-4x+3)
=x^2(x-3)(x-9)+27(x-3)(x-1)=(x-3){x^2(x-9)+27(x-1)}=(x-3)(x^3-9x^2+27x-27)
Apply the fomula: (a-b)^3=a^3-3(a^2)b+3ab^2-b^3 for factoring x^3-9x^2+27x-27)
we have: a=1 and b=3, and x^3-12x^2+27x-27=(x-3)^3, so g(x)=(x-3)(x-3)^3=(x-3)^4
Thus, f(x)=xg(x)=x(x-3)^4. That is: x^5-12x^4+54x^3-108x^2+81x=x(x-3)^4=0
We have: x=0 or x=3
CK: Plug the values of x into Eq.1. We have: f(0)=0 and f(3)=0 CKD.
Therefore, the answers are: x=0 or x=3