Question 257025
The original question was
f(x) =(x^2-4)^3.
step 1 - set = 0 to get
(i) {{{f(x) =(x^2-4)^3 = 0}}}
take a cube root to get
(ii) {{{f(x) =(x^2-4) = 0}}}
Factoring, we get
(iii) {{{(x-2)(x+2) = 0}}}
x = 2 and x = -2
It turns out that 
x = 2 with multiplicity 3
x = -2 with multiplicity 3