n(n-2)(n+2)=n^3-4n

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Solution for n(n-2)(n+2)=n^3-4n equation: 


Simplifying
n(n + -2)(n + 2) = n3 + -4n

Reorder the terms:
n(-2 + n)(n + 2) = n3 + -4n

Reorder the terms:
n(-2 + n)(2 + n) = n3 + -4n

Multiply (-2 + n) * (2 + n)
n(-2(2 + n) + n(2 + n)) = n3 + -4n
n((2 * -2 + n * -2) + n(2 + n)) = n3 + -4n
n((-4 + -2n) + n(2 + n)) = n3 + -4n
n(-4 + -2n + (2 * n + n * n)) = n3 + -4n
n(-4 + -2n + (2n + n2)) = n3 + -4n

Combine like terms: -2n + 2n = 0
n(-4 + 0 + n2) = n3 + -4n
n(-4 + n2) = n3 + -4n
(-4 * n + n2 * n) = n3 + -4n
(-4n + n3) = n3 + -4n

Reorder the terms:
-4n + n3 = -4n + n3

Add '4n' to each side of the equation.
-4n + 4n + n3 = -4n + 4n + n3

Combine like terms: -4n + 4n = 0
0 + n3 = -4n + 4n + n3
n3 = -4n + 4n + n3

Combine like terms: -4n + 4n = 0
n3 = 0 + n3
n3 = n3

Add '-1n3' to each side of the equation.
n3 + -1n3 = n3 + -1n3

Combine like terms: n3 + -1n3 = 0
0 = n3 + -1n3

Combine like terms: n3 + -1n3 = 0
0 = 0

Solving
0 = 0

Couldn't find a variable to solve for.

This equation is an identity, all real numbers are solutions.

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