How do you solve $0 = - n^{3} + n$ $0= -n^3 + n$ ?

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Answer 1 · 2017-02-14T15:40:45

$n = 0, n = - 1, n = 1$ $n=0," " n=-1," " n=1$

Let's change the equation around a but first:

$n^{3} - n = 0 \leftarrow$ $n^3 -n = 0" "larr$ first term is positive.

$n (n^{2} - 1) = 0$ $n(n^2-1) =0" "$ factorise with a common factor

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

Each factor can be equal to 0.

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

Solutions are:

$n = 0, n = - 1, n = 1$ $n=0," " n=-1," " n=1$

How do you solve 0=−n3+n 0= -n^3 + n?