Let $f(x) = ax^6 + bx^4 - cx^2 + 3.$  If $f(91) = 1$, find $f(91) + f(-91)$.
So we know that $f(x) = ax^6 + bx^4 - cx^2 + 3$, and we want to find $f(91) + f(-91)$. Right, and we already know that $f(91)$ is 1. Let's start by finding $f(-91)$. We know that $f(x) = ax^6 + bx^4 - cx^2 + 3$, so to find $f(-91)$ we just need to replace $x$ with $-91$. And if we do that we get $a(-91)^6 + b(-91)^4 - c(-91)^2 + 3$. Right. So $f(-91) = a(-91)^6 + b(-91)^4 - c(-91)^2 + 3$. Wait a minute. If we look closely we can see that $(-91)^6 = 91^6$ and $(-91)^4 = 91^4$. Right. $(-91)^6 = (-1)^6 * 91^6 = 1 * 91^6 = 91^6$. And $(-91)^4 = (-1)^4 * 91^4 = 1 * 91^4 = 91^4$. So $f(-91) = a(-91)^6 + b(-91)^4 - c(-91)^2 + 3$ is actually $f(-91) = a(91)^6 + b(91)^4 - c(91)^2 + 3$. Now we know that $f(91) = 1$, so we can write the equation $1 = a(91)^6 + b(91)^4 - c(91)^2 + 3$. And we want to find $f(91) + f(-91)$ which is $1 + f(-91)$, so we can write the equation $1 + f(-91) = a(91)^6 + b(91)^4 - c(91)^2 + 3 + a(91)^6 + b(91)^4 - c(91)^2 + 3$. If we simplify that we get $1 + f(-91) = 2a(91)^6 + 2b(91)^4 - 2c(91)^2 + 6$. And if we substitute $1$ for $f(91)$ we get $2 = 2a(91)^6 + 2b(91)^4 - 2c(91)^2 + 6$. And since $f(91)=1$, we know $f(-91)=1$. So $1 + f(-91) = 1 + 1 = 2$. Right!

