Solve to find the unknown I have been doing questions from the past year and I come across this question which stumped me:

The constant term in the expansion of $\left(\frac1{x^2}+ax\right)^6$
  is $1215$; find the value of $a$. (The given answer is: $\pm 3$ )

Should be an easy one, but I don't know how to begin. Some help please?

Via the binomial theorem we know the $x^0$ factor (the coefficient of which you call the "constant term") is actually
$${6 \choose 2}\left(\frac{1}{x^2}\right)^2(ax)^{6-2}.$$
Since 6 choose 2 is equal to $5\cdot6/2=15$, we have the equality $15a^4=1215$. The prime factorization of 1215 is $5\cdot3^5$, so we arrive at the conclusion that $a^4=3^4$, or $a=\pm3$ (discarding imaginary solutions). 
