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

Question

Ethan Adkins · Accepted Answer

#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#

Jameson Allen · Answer

undefined To solve the equation $0 = -n^3 + n$, follow these steps:

1. Factor out the common factor $n$: 
$$0 = n(-n^2 + 1)$$

2. Set each factor equal to zero and solve for $n$:
$$n = 0$$ or $$-n^2 + 1 = 0$$

3. For $-n^2 + 1 = 0$, isolate $n^2$:
$$n^2 = 1$$

4. Take the square root of both sides:
$$n = \pm \sqrt{1}$$

5. Simplify the square root:
$$n = \pm 1$$

So, the solutions to the equation $0 = -n^3 + n$ are $n = 0$, $n = 1$, and $n = -1$.

Avery Alexander · Answer

undefined To solve the equation $0 = -n^3 + n$, you first factor out $n$ to get $n(-n^2 + 1)$. Then, factor the quadratic expression $-n^2 + 1$ as $-(n^2 - 1)$. Finally, factor the expression $n^2 - 1$ as $(n - 1)(n + 1)$. So, the factored form of the equation becomes $n(n - 1)(n + 1) = 0$. Now, since the product of factors equals zero, one or more of the factors must be zero. Thus, the solutions are $n = 0$, $n = 1$, and $n = -1$.