how do u find the root of

m^3+3m^2+3m+1 = 0

please help me find what are the value of what m is equal to

2 answers

assistance needed
I would let f(m) = m^3+3m^2+3m+1

on second try f(-1) = 0, so m+1 is a factor
using sysnthetic division, or long algebraic division, the other factor is (m^2 + 2m + 1)
which is (m+1)(m+1)

so m^3+3m^2+3m+1 = 0
(m+1)^3 = 0
m = -1

(I should have recognized 1 3 3 1 as the third row of Pascal's triangle)
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