Sunil is asked for the value of ${\sin }^{6}A+{\cos }^{6}A+3{\sin }^{2}A{\cos }^{2}A$ for ANY angle A.
What single number should he write?

Put $x={\sin }^{2}A$ and $y={\cos }^{2}A$, so $x+y=1$. Then $x^3+y^3=(x+y{)}^3-3xy(x+y)=1-3xy$, and adding $3xy$ gives 1. So the expression equals 1 for every A. Answering 3 comes from mis-expanding the cube and forgetting that $x+y=1$ makes the leftover terms cancel exactly.