Given n is a root of quadratic equation $x^2+5x-12=0$. Show that $n^3=37n-60$. Does this question have any trick or require any special mathematical skill?
It's just a little trick:
That is, substitute $n^2$ each time it occurs.
Another perspective is this:
If you start with $n^3-37n+60$ you can divide by $n^2+5n-12$, this will be a factor (looking at the coefficient of $n^3$ and the constant term) if $$n^3-37n+60=(n-5)(n^2+5n-12)$$ It is straightforward to check this is true (or you could use polynomial division rather than spotting a factor).
Once you have this identity it is trivial that if the right-hand side is zero, so is the left-hand side.
Note that the solutions starting with $n^3$ and reducing the power using the equation given, provide an easy way of showing that if $n$ satisfies a quadratic equation (in suitable circumstances), any power of $n$ can be expressed as a linear expression in $n$. And this can be generalised to show that if $n$ satisfies a polynomial of degree $r$ and polynomial expression in $n$ can be reduced to an expression where the highest power of $n$ is less than $r$.
Suitable circumstances above means that the polynomial should be monic, or should be reducible to monic by dividing through by a constant factor. This is important, for example, when it comes to working with algebraic integers.
The same reduction can be achieved using the division algorithm.
The method I have used would not necessarily work if the original polynomial had a double root. If I started with $p(x)=(x-2)^2$, I could say if $p(x)=0$ so is $q(x)=(x-5)(x-2)$ - but here $p(x)$ is not a factor of $q(x)$. But other methods can also fail at this point. It is easy to identify repeated factors/double roots in specific cases. The general case (to show it always works) requires some algebraic machinery.
No it's pretty straight forward.
If $n$ is a solution then $n^2 + 5n -12 = 0$ and so $n^2 =12 -5n$, Now multiply each side by $n$ you get $$n^3 = 12n -5n^2 = 12n -5(12-5n) = 37 - 60n$$