 Hi, I'm Zor. Welcome to Unizor Education. I'd like to solve a small problem right now. The problem is about proving an identity of some algebraic expressions. As you know, we're talking about equations in general, and to solve equations, you have to basically transform algebraic expression from one form to another. And the skill of transformation is actually very important, it gives you the possibility to solve equations which you might not even know like a general solution or a formula about. So actually this problem illustrates the main purpose of all the lectures which we are trying to arrange here right now because it's not about applying certain things we should know, it's about trying to walk the way which you never walked before to come up with certain, maybe a little trick, maybe a little guess which will allow you to solve something which you think is unsolvable basically. And here it is basically. Let's consider that you would like to solve an equation of a really complex type and I will just put it here. Now, obviously if you want to solve this equation, maybe another one, why maybe some number, whatever the number is, it will be equation for the x to the power 4 which obviously is very difficult to solve. However, if you will be able to transform it into a nicer form which is easily to solve, that will be the way. So the question is how to transform one equation into another. So there are a couple of, I mean not a couple actually, a lot of problems which are dedicated to transformation from one algebraic expression to another one which is identical but it's just different. And the second form is much easier to solve. So what I would like to do right now, I would like to prove the identity of this expression on the left part of the equation to this. Now, why is it better? Again, let's consider y is a number. Instead of solving an equation that this is equal to zero, you can solve an equation that this is equal to zero. Now, this is equal to zero, it's square, so basically it means that everything inside the parenthesis is equal to zero, square or no square. And this is the quadratic equation for x. So basically to solve this equation of the fourth degree, you can solve the square, the quadratic equation and that would be much simpler obviously. So you reduce the complexity of the problem. All right, so now, again, the problem is to prove this identity. These two expressions are exactly the same. Now, well, I should say the easy way but straightforward way of doing this is x plus four multiplied by itself four times x plus y times x plus y times x plus y times x plus y have a big formula and then do exactly the same thing, multiply this expression by itself to make it square and then if you will compare left and right sides of this expression, you will see that they are identical. So that's something which again, I would say, not the easy but the straightforward way. Yes, you can do it, but there is no fun in this, do with me. So let's try to do something, maybe a little trick, a little guess and if you never guessed anything about this, you might actually have a problem. I mean by all means try to do it yourself. However, if you will solve certain number of equations, in certain cases you will maybe listen to somebody else who helped you, in certain cases you will come up with your own solutions. So basically you will develop a skill of guessing, so to speak. So one of the, I don't know, you can call it a trick or some kind of approach which you might take. You see these guys are quite symmetrical and you will see that there are two very important expressions here, x plus y and xy and since they are symmetrical, here is something which will definitely help to simplify proving of these two things and here is what I mean. If I will substitute instead of x and y, I will use different variable. I will use variable u which is x plus y and v which is x times y. I would like to express both sides in terms of u and v and in terms of u and v it will be much simpler to prove. All right, well let's try. On the left, x plus y to the force this means u to the force, u to the force degree. Now, x to the force and y to the force, well that's too much actually. However, what we can do is the following. We can have x to the force plus 2x squared y squared plus y to the force. Now, do you recognize this? Well, it has x to the force and y to the force and we have added this piece which is expressible in terms of u and v obviously. But is this easier? Well, let's think about this. Obviously, this is x squared plus y squared, right? x to the force times double product times y to the force and this is closer to the expressions which we have because you can have it again inside. We will do exactly the same trick. We will do plus 2x and minus 2x so it will be x squared plus 2x y and minus 2x y plus y squared. And this piece, again you can recognize this is x minus, sorry, not this one. Yes, this and this and this. This is x plus y squared, right? These three numbers. So, this is x plus y squared minus xy and the whole thing squared still remains. So, what I would like to say is that this expression we have transformed into this, this into this and this into this. And here we have only x plus y and xy which are our new variables. So, if I want only x fourth and y fourth, these two guys, I have to use this without this plus. So, this is u squared minus 2y to xy which is 2v squared. This is the whole expression but this is definitely something which I could add an extra and I have to subtract it back now which is 2 and xy squared is v squared. That's my left part. Well, I think it is easier and you will see why. Now, the right part also can be transformed into u and v how? By adding xy. So, you will have 2 times x squared plus 2xy plus y squared and minus xy squared. Why do they do this? 2xy minus xy, that's 1xy. Because again, this is x plus y squared. So, again I am expressing my expression in x and y in terms of u and v. So, what I can say is that this is equal to 2 open parenthesis u squared minus v. That's the right part. So, let's write it here. Oh, square, I'm sorry. So, I have to prove that this is 2u squared minus v squared. So, what I did, this is my first step. I have re-signed the variables x and v and expressed them in terms of u and v and the whole x and y in terms of u and v. And the whole expression in terms of x and y, I have transformed into expression over u and v. And quite frankly, this even looks a little simpler. Because I don't have something to the fourth degree which I have to multiply by itself and produce God knows how many members in this huge sum. The most complex thing which I have here is I have two expressions which are just single by themselves squared, which is kind of easy thing. Now I can open these parenthesis, I can square these two expressions and what will I have u to the fourth plus? Well, this is just playing the difference between two variables squared. So, the square of the first one which is u to the fourth minus double that product which is 4u squared v plus square of the second one which is plus 4v squared minus 2v squared. That's on the left. And on the right, I have to open this square which is 2u to the fourth minus 2u squared v plus v squared. Is this a true identity? Well, let's check it out. Let's open these parenthesis. So, 2 times u to the fourth minus 4 and plus 2v squared. Let's check it out. u to the fourth to u to the fourth. That's fine. Minus 4u squared v minus 4u squared v. 4v squared minus 2 and this is playing 2 again. So, identity is proven in terms of u and v. So, basically what I would like to say is that if I started doing everything in terms of x and y I would have to multiply x plus y to the fourth degree which is the second degree, it will be trinomial and then trinomial by itself, it will be nine members and this is another trinomial, another nine members so it will be 18 different members in my expression and I'm not saying it will be like significantly difficult but it's really definitely tedious. This is a nice, I would say, approach which you can take. Substitution, something which you think might simplify in the future your life and in this particular case I have just used this particular approach. I substituted this expression, I expressed it in terms of u and v and this one as well and in terms of u and v it looks a little simpler to prove and basically that's it. Is it the only way to prove this identity? Absolutely not, I'm sure there are some other ways, I just didn't come up with anything but this is just one of the ways and why is it important? Well, you remember we will have quadratic equations and in the solving quadratic equations you will see that each of two solutions and product of two solutions play a very important role in the quadratic equations. The product is equal to the free member of the quadratic equation and the sum is equal to the second coefficient with a negative sign. You will see it in further lectures on quadratic equations so sum and product are very often occurring in algebraic equations etc. So if you see that something is really symmetrical relative to sum and product of two members of two unknowns it might actually very well make sense to consider substitution, certain more complex expressions with something else which is expressed in sum and product of two different unknowns. So just one of the approach, remember sometimes it might be helpful and what's more important in this particular case is to realize that even that something is possible to do straightforward in certain cases it might make sense to think about a simpler approach. The question is what's the approach and whether you will be able to come up with this approach? Well, it depends. The more problems like this you will solve yourself. The more equipped you will be even in real life to find a simpler solution for something which straightforward might not be really doable basically at all. So thanks very much for listening with me. Don't forget to check the Unisor.com website. There are many other aspects of mathematics over there and it might be interesting for you. Thanks.