## Precalculus (6th Edition) Blitzer

$16$, $1$, $128$.
We can rewrite the equation as $9(x^2-8x+16)-16(y^2+2y+1)=16+9(16)-16$ which means that we need to add $16$ within the first parentheses to complete the square on $x$, and add $1$ to complete the square on $y$. The amount we need to add on the right side is $144-16=128$.