 I'm Zor. Welcome to Unizor education. Today we will just solve a very simple but probably illustrative problem about equations and transformation basically from one expression to another. In this case we will use something which is called conditional transformations. Okay, so here is a problem, very simple one. We will basically solve it together. And here it is. Let's say it's given that three numbers satisfy this particular condition. What has to be proven is the following identity. That a cubed plus b cubed plus c cubed equals 3abc. Now obviously for a, a, b and c this is not really an identity because it definitely depends on how, what's the values are. However, if this is a condition under this condition, this particular expression is an identity. Now there are certain unconditional identities. Well, for instance, like something like x plus y square equals to x square plus 2xy plus y square. I mean this is something which is unconditional because if you will multiply any sum of any two numbers by itself and open all the parentheses and gather together similar members you will get exactly this. So this is unconditional identity. This identity is conditional and this is a condition. All right, so let's try to prove it. Well, the simplest way to prove this particular conditional identity is just to use this, let's say to resolve it for a for instance and substitute it into something which we would like to prove. All right, so getting a from this will be a equals to minus b plus c because this is identically transformed into this. Now we know that a is expressed in terms of b and c using this particular formula. So let's substitute it here and let's see what happens. Well, a cube, let's do it in steps. First we will do b plus c square which is equal to b square plus 2bc plus c square in this parenthesis. So b plus c cube is equal to b plus c times this expression b square plus 2bc plus c square which is equal to, let's multiply b by any of these, open the parenthesis. It will be b cube plus 2b times 2bc which is b square c and b times c square plus bc square. Now c multiplied by any of these guys, it will be b square c, let's use alphabetical order of letters, 2b square, 2bc square and the last one will be c cube. Okay, so this is b plus c cube, well, which is different from a cube just by the sign. So if I want to have a cube, that would be this expression with a minus sign. Since I don't want to rewrite everything with a minus sign, this is equal to minus a cube. So that's something which I have obtained only using our condition. Now let's think about how can I use it from here. Alright, now instead of, since I have minus a cube here, instead of proving this particular identity, I will transfer a to the right side and I will have b cube plus c cube equals to 3abc minus a cube. So I will prove this particular identity instead of this because these are identical to each other. Alright, so how can I prove this one? Well, I have an expression for minus a cube and I also have an expression for a. a is equal to, as you remember, minus b plus c. So I'll just substitute. Instead of minus a cube, I will substitute this. Instead of a, I will substitute this. Well, and let's see what happens. By the way, again, instead of worrying about the sign, it would be easier if I will write it down in the following fashion. Minus a equals b plus c. That's the same thing, right? And here I will do something similar. I will transfer 3abc on the left part. So it will be minus 3abc. And on the right, I will have minus a cube. Again, everything is completely identical. I transfer a cube to the right, 3abc to the left by subtracting a cube and subtracting 3abc from both sides. So I will prove this one and it's easier because there is already a sign minus and sign minus here. So I don't have to worry about signs. It will always be plus. So let's substitute. Instead of minus a, I will substitute it here, b plus c. It will be b cube plus c cube. Now minus 3abc, it will be plus 3b plus c. That's a, right? Instead of minus a, I put plus 3b plus c times bc. And instead of minus i cube, I have to substitute this expression. So now I have this as an identity which I have to prove, or to check, actually. Well, actually, yes. It's better just to check because it's completely identical to pieces. We just have to very carefully look at this. Alright. Why are they identical? Well, B-cube and B-cube. Good. C-cube and C-cube. So far so good. Now this is 3B square C. Let me substitute it. 3B square C. If I open these parentheses and multiply by BC, it will be B-square C plus 3BC square. Now B-square C once and B-square C twice. So this plus this, it's 3B square C. And this is 3B square C. So these go out. And similar to this, 3BC square and here I have BC square plus 2B square. Again, these are out. So everything is reduced. You see, all members on the left and all members on the right are equal to each other. So basically that's it. I have proven that this particular thing is identity provided this particular condition. Well, that's it. It's a very easy thing. Just exercise for careful substitution and multiplication of members of these identities. And I will probably put some more problems of this type for your self-study. Try to do everything whatever it is, even if there is no solution like this one, which I have provided. This is just an illustrative example how to deal with conditional identities. So please do as much as you can. And thanks for your attention today. Don't forget to check for Unisor.com. There are many other interesting things related to mathematics on this side. And you're welcome to basically take a look at them. Thanks very much.