How do you solve # 0= -n^3 + n#?

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Answer 1 · 2017-02-14T15:40:45

#n=0," " n=-1," " n=1#

Let's change the equation around a but first:

#n^3 -n = 0" "larr# first term is positive.

#n(n^2-1) =0" "# factorise with a common factor

#n(n+1)(n-1) =0" "# difference of squares

Each factor can be equal to 0.

#n=0, " " n+1=0" "n-1=0#

Solutions are:

#n=0," " n=-1," " n=1#