 I am Zor. Welcome to Unizor Education. Today's topic will be another problem in the area of equations. As I have explained before, the purpose of all these lectures is not just to give you some formulas and you can apply them in real practical situations, but to put something in front of you which will force you to think about certain problems in certain non-trivial ways. So, I am trying to come up with problems which will really encourage you to think about these things non-trivial. Okay, so today's equation will be, and that's the problem number 4, x to the fourth plus, oh sorry, that's too much, x to the third plus x square plus x plus 1 equals to 0. Okay, now this is a cubic equation, the equation of power of 3, and there is no simple formula to generally solve this particular equation just for any kind of coefficients with x cube, x square, x and 1. So, I am not asking you to just solve any cubicle equation, I am asking you to solve this particular one. And considering nobody actually taught you about how to solve generally all the cubicle equations, so this particular equation should have certain specifics which you have to recognize and use to solve the equation. And that's exactly what I am asking you to do right now. Think about this, look at some specifics which might really help you to solve this particular equation and after you find them, use these things to simplify it and finally come up with a solution. And this is obviously a good time to press the pause button, think about this particular equation and I will continue meanwhile to basically explain how to do it. Well, let me just start with one particular way to solve this equation. I am just looking at this and I recognize that if I will, oh one more thing before I start it, any equation should have the domain where I am looking the solutions for. In this particular case, you know the typical domains are real numbers and complex numbers, so I will try to basically approach both. Alright, so let's start now just looking at this equation. I can recognize the fact that if I will factor out x squared from these two guys, I will get this. Right, x squared times x is x cubed, x squared plus 1 times 1 is x squared. And this also x plus 1. Well, obviously the next step would be to factor out x plus 1 and I will have x squared plus 1, x squared plus 1 equals to 0. Obviously, if you have a product of two expressions which contain unknown x is equal to 0, then obviously one solution is when this equation is equal to 0, I mean this expression is equal to 0 and another solution is when this is equal to 0. So we have already from one particular equation, we have two different equations, x plus 1 equals to 0 and x squared plus 1 equals to 0. Any solution to this equation will be solution to this one and any equation of this, any solution to this equation will be solution to this one because if this particular expression is 0, then the whole product is 0 and same with this one. Alright, next step obviously is just listing the solutions. This is one solution and this is good at both domains in all real numbers and in all complex numbers because it's real. Now as far as this is concerned, obviously x squared equals to minus 1 has no real solutions but it has solutions among the complex numbers and these are x equals to i and x equals to minus i. So these two are additional solutions in the domain of complex numbers. But among real numbers we have one solution only to this original equation and among complex numbers we have three solutions, this one plus two additional, plus minus i. Well, just out of curiosity, let's check these solutions. By the way, all these transformations are invariant which means we have not lost anything. So exactly three solutions in the area of complex numbers, any equation of power 3 should have and we have it. Alright, so let's just check one of them, the most difficult one if you wish, x equals to minus i. Alright, let's do this. Minus i to the third degree plus minus i squared plus minus i plus 1 equals to zero. Alright, now minus i to the third degree is minus 1 to the third and i to the third, right? Because minus i is minus 1 times i so i is separate and minus 1 is separate. Now minus 1 to the third degree, power 3 gives you minus 1 times minus 1 times minus 1 which is minus and i cubed is actually i squared times i. Now i squared as we know is minus 1. So minus minus and this minus and this i will give us just plane i. Now i squared or minus i squared is minus 1. So this is minus 1. This is minus i and this is plus 1. And as you see, this is reducible and this is reducible equals to zero. The solution is correct which I actually never doubted because we had all the transformations inherent in this particular case. So that's good. Now that's one way of solving this particular equation and let's just go back to the original way of doing this. If you remember, you looked at this equation and it just come up to us that if you factor out x square you will get x plus 1 and this x plus 1 so you can factor x plus 1 again. Now there are different solutions to any equation. So this particular one has yet another solution which I would like to present you. Which also is if you wish some kind of a trick. Anything which requires some idea but in this particular case factor out x square and then factor out x plus 1. Anything non-standard requires some kind of idea because the standard ones you just use the formula and that's it. But that's not what we are talking about in this particular case. We are talking about certain non-trivial solutions in case there are no trivial, basically there is no trivial solution to this equation. So we need to find out something special, some trick. Okay, so one trick was factoring out x square and x plus 1. Here is another trick. Let's multiply this equation by x minus 1. But if I multiply x minus 1 to the right, obviously 0 will remain 0. Now, first of all you have to understand that this is not an invariant transformation because I am multiplying something which contains an unknown. So if I am multiplying by something which contains unknown if that particular something is equal to 0 then I am basically breaking the rules of invariant transformations. We know that invariant transformation is when both sides of the equation can be multiplied by a non-zero factor. In this case I don't know what I am multiplying by. So first of all I have to say when I am multiplying by x minus 1 that x equals to 1 might not. Actually, let's express a different thing. x should not be equal to minus 1 when I am multiplying by x minus 1. Sorry, 1 minus 1. So this is a condition. But my original question is yes, I am excluding x equals to 1 from the solutions of the original equation because I am multiplying by x minus 1 which is supposed to be not equal to 0. Question is if I will put this condition will I lose anything among solutions? Well, let's just check. Is 1 a solution? No, because 1 plus 1 plus 1 plus 1 is 4. So 1 is not a solution. So this particular condition which I am imposing right now doesn't really reduce the set of solutions which we have. We just checked it, it's not a solution. And now even if after I will solve this equation if I will get the solution x equals to 1 I should really not use it. Okay, so that's how I deal with non-invariant transformation. First I exclude the case when the whole expression is equal to 0 which means x equals to 1. And while I am excluding this I am checking that 1 is not really a solution. So I am not losing anything and everything is great. Now I can consider the resulting equation in the area of x not equal to 1. Alright. Now, why did they multiply it by x minus 1? Okay, here is the work. Let's just open these parenthesis. x times each one of those will be x to the fourth plus x to the third plus x to the second and x. Now if I multiply minus 1 by these guys I will get minus x to the third minus x to the second minus x and minus 1. That would be when I multiply minus 1 by these guys, right? And this is equal to 0. Now obviously this is reducible and what do I have as a result? I have x4 minus 1. So x4 minus 1 equals to 0. x4 is equal to 0. So this is why I have multiplied by x minus 1 to get this equation. Now you can ask why did I complicate my life transforming the equation of the power of 3 to the equation of the power of 4? Because this is much simpler, obviously. Now how to solve this equation? This is really elementary. So x to the fourth equals to 1. Now, very carefully we can extract the square root from both sides. In this particular case, square root from both sides is x square equals to 1. Now you remember that when I was extracting the square root I have to use absolute values in both sides because I need the positive numbers. Well, x square is always non-negative so I can just drop the absolute value and one is non-negative as well. From here I get here, or I can put it, I can drop it, it doesn't really matter. So the square root of 1 is absolute value of 1. But let's just do it again a little more clearly. I did it too fast. Square root of x to the fourth is x square which is always non-negative. And this one is absolute value of 1 which means that x square is equal to 1 is one solution and x square is equal to minus 1 is another solution. So that's simple. We transformed our original equation to two very simple equations. Now, this is the same thing again. So x is equal to 1 and x is equal to minus 1. Two solutions. Now, this in the area of real numbers doesn't have any solutions. In the area of complex numbers it's again something which we are familiar with. And now we have four solutions. Well, if you remember the first attempt, the first method of solving this particular equation gave us only three solutions. It was i and minus i and minus 1. This gives us four. But do you remember when I first multiplied the whole thing by x minus 1, I said x should not be equal to 1. So this is not a solution. Now, I'm not saying that the second method is simpler than the first one. Actually, maybe calculations are a little more trivial in this case. However, what's less trivial is the fact that we have used non-invary transformations. And we have to basically analyze this x equals to 1 as an extraneous solution which we should really disregard. But in any case, I just wanted to present these two methods because sometimes one is working, sometimes another is working. It all depends on your taste and specifics of a particular cast. But this is actually a very interesting example of using the formula which simplifies lots of different things. So let's finish up with this and let me give you a formula which basically kind of generalizes this particular equation. So if I have x to the power of n minus 1 and minus 2, this expression, if I will multiply it by x minus 1, well, it will be exactly the same thing as in the case of the cubicle equation. So if I multiply x by any one of these guys, I will get x to the m plus x to the m minus 1 plus, et cetera, plus x to the third x to the square plus x. Then minus 1 minus 1 times x to the m minus 1 minus, et cetera, minus x cubed minus x to the square minus x minus 1. That's what will happen. And obviously all these will reduce and the result will be x to the m minus 1. So this is a formula which you probably just, you know, have to have as your one of the methodology to solve certain equations. To use this formula, many kids just have to remember it probably because there is nothing more interesting. Just remember that this formula actually does simplify in some way certain equations. Well, basically that's all I wanted to talk about today. These two methods to solve this particular cubicle equation and thanks for your attention and don't forget that lots of different things you can find at unison.com. That's mathematics for teenagers. Thank you very much.