What is the answer to ((n+1)!)/((n-2)!)=n^3-n ?

The answer to ((n+1)!)/((n-2)!)=n^3-n is True for all n

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((n+1)!)/((n-2)!)=n^3-n

\frac{( n + 1 ) !}{( n - 2 ) !} = n^{3} - n

True for all n

Solution steps

\frac{( n + 1 ) !}{( n - 2 ) !} = n^{3} - n

Simplify

\frac{( n + 1 ) !}{( n - 2 ) !} : n (n + 1) (n - 1)

n (n + 1) (n - 1) = n^{3} - n

Solve

n (n + 1) (n - 1) = n^{3} - n :

True for all

n

True for all n

5 x^{- 2} - 9 x^{- 1} - 2 = 0

a, b + \frac{m}{a} = z

\frac{18}{7 + c} = \frac{10}{7 - c}

\frac{6}{2 x + 7} = \frac{10}{3 x - 8}

E, L = \frac{25 M}{E}

What is the answer to ((n+1)!)/((n-2)!)=n^3-n ?
The answer to ((n+1)!)/((n-2)!)=n^3-n is True for all n
Solve for n: ((n+1)!)/((n-2)!)=n^3-n
The solution is True for all n