Solve an equation involving factorial: $\frac{(n+1)!}{(n-2)!}=990$

Question

For this equation:

$$\frac{(n+1)!}{(n-2)!}=990$$

I need help with the working to the answer. Well I was stuck on the bit where I had ended up with: $$(n+1)n(n-1)=990$$ $$(n^2-1)n=990$$ $$n^3-n=990$$

Answer 1

$$\frac{(n+1)!}{(n-2)!} = 990$$ $$ \frac{(n+1)(n)(n-1)(n-2)!}{(n-2)!} = 990$$ $$(n+1)(n)(n-1) = 990$$ $$(n+1)(n)(n-1) = 99 \times 10$$ $$(n+1)(n)(n-1) = 11 \times 10 \times 9$$

thus $n=10$

Answer 2

Hint: Cancellation by division of factorial common factors in numerator and denominator should leave you with three terms, giving a cubic polynomial.