For what value of $n$ is $3^3-5=4^2+n$?
To solve for $n$, I need to isolate it on one side of the equation by using inverse operations. First, I can subtract $4^2$ from both sides to get $3^3-5-4^2=n$. Next, I can simplify the expression on the left by evaluating the exponents and the subtraction. I know that $3^3=27$, $4^2=16$, and $27-5-16=6$. So, the equation becomes $6=n$. This means that $n=6$ is the value that makes the equation true. 