 In this video, we will present the solution to question number 15 for practice exam number one for math 1210. We're given a function f of x equals x cubed, and we're supposed to evaluate and simplify the difference quotient f of x plus h minus f of x over h. So as we're computing and simplifying this difference quotient, we need to plug in to f of x, the x plus h. That's a very important part. Many students make the mistake of thinking that f of x plus h is equal to f of x plus h, which when you say it out loud in words, it makes sense, but the two things are very, very different. What we need to do is we need to replace every x in f of x with an x plus h. So just a racing real quick. We end up with the following. We're going to get an x plus h cubed. We didn't have to subtract it. So this right here is the x, the f of x plus h. We replaced the x with an x plus h. Then we're going to f of x. That's just the original function again, which is x cubed, and then the denominator is h. Our goal is to get rid of the h in the denominator. Now for this one, how we proceed depends on the function type as this is a x cubed. We're going to have to foil that thing out. Using the binomial theorem, right, we can reconstruct Pascal's triangle here. And so when you take x plus h cubed, the coefficients are going to be 1, 3, 3, 1. And so this would expand to be x cubed plus 3x squared h plus 3x h squared plus h cubed minus x cubed all over h. For which then you're going to see that the x cubes will cancel out. Everything from the f of x part will cancel with something from the f of x plus h part. So you're left with 3x squared h plus 3x h squared plus h cubed. You'll notice that after we canceled out the f of x part, everyone in the numerator is now divisible by h. We factor that out. So we get 3x squared plus 3x h plus h squared all over h. And then the h in the denominator will cancel with that multiple of h in the numerator. And so we see that the simplified difference quotient will be 3x squared plus 3x h plus h squared. And so this right here is the simplified difference quotient. There's nothing more to do. We don't do anything else because we're not calculating the derivative in this example. We're just asked to simplify the difference quotient.