Basic Factorial Ratio issue I was reading the IB mathematics book and I came across an equation related to factorial and ratios. The equation is 
$$\frac{(n+1)!}{(n-2)!} = 990$$
so these types of equations generally lead to a quadratic of some sort however this one leads to $n^3 - n = 990$. I have no idea how to solve this, the answer is $10$ but I have no clue how I end up with that. I can obviously guess that from looking at the equation, but what is the correct way to do it?
*it is stated in the answer sheet that quadratic is the way to go but I can't really figure out how
entire process of what i do is : 
$$(n+1)n(n-1)(n-2)! / (n-2)! = 990 ~> (n+1)n(n-1) = 990$$
$$(n^2+n)(n-1) = 990 ~> n^3 + n^2 - n^2 - n = 990 ~> n^3 - n = 990$$
tried doing: $n (n^2-1) = 900$ ~> $n^2 - 1 = 900/n$ ~> $n^2 - 900/n -1$ but didn't really help me.
Thank you for your response.
 A: Since $n$ is an integer $\ge2$ and $n^3-n$ is strictly increasing on that range, there are just finitely many possible cases to check. Specifically, you can bound them by observing that the condition $n-1\le \lfloor\sqrt[3]{990}\rfloor=9$ is necessary.
A: $$\dfrac{(n+1)!}{(n-2)!}=\dfrac{(n+1)n(n-1)(n-2)!}{(n-2)!}=(n+1)n(n-1)$$
From here we get $n^3-n=990$
Now... what you could do from here is apply cardano's method which will eventually give you the solution... however it is incredibly tedious and it is incredibly rare for anyone to actually memorize it.
Instead, we can recognize that $n^3-n$ is approximately equal to $n$ for large enough $n$, and so $n^3$ should be approximately equal to $990$.
Taking the cubic root of both sides should give us then a good approximation., which gives us $\sqrt[3]{n^3-n}\approx\sqrt[3]{n^3}=n\approx \sqrt[3]{990}\approx 9.967$
So, we learn that $n\approx 9.967$.  Since the solution to a problem which is written well of this type should have it be that $n$ is a natural number, the closest natural number to this is $10$.
Indeed, plugging it in, we see that $(10+1)\cdot 10\cdot (10-1)=11\cdot 10\cdot 9 = 990$ and so $n=10$ is the solution.

Note: The first method might make you think that $n=10$ would be the solution for $\dfrac{(n+1)!}{(n-2)!}=995$ as well using the same method.  It is for this reason that the double-check at the end is important and with the right argument you will be able to show that $\dfrac{(n+1)!}{(n-2)!}=995$ is impossible for $n$ a natural number (e.g. that $\frac{(n+1)!}{(n-2)!}$ is always even).  Cardano's method would be able to find the real number $x$ such that $x^3-x=995$ though even if $x$ turns out not to be a whole number.

Another option would be to look at factorization of $990$ which you should be able to see $990=10\cdot 99 = 10\cdot 9\cdot 11$ and you could factor further, but you should notice that this should scream out the solution as well.
