Many get stuck on this exponential algebra math equation.

Can we find the value of x? Yes, we can. Welcome back to my channel. This looks complicated, but we can find a way around it as usual, right? So, today we are asked to find the value of x if x cubed plus 1 over x cubed equals 110. So first, let's note that we can write this as x cubed, like this, plus 1 over x, all raised to the power of 3 as well, and this equals 110. Now, recall the algebraic identity, right, that states that a cubed plus b cubed equals a plus b times a squared minus ab plus b squared, right? In this case, our a is x, while our b is 1 over x. So, our a is x, our b is 1 over x. So, we're going to have a plus b, which is x, plus 1 over x, multiplied by a squared, which is x squared, minus ab, which is x multiplied by 1 over x, plus b squared, 1 over x squared, right? And everything is 110. If this is the first time you've seen this lovely, wonderful channel, please hit that subscription button and turn on your notification bell, yes. If not, this might be the last time you'll see me, okay, and I wouldn't want to miss you in my next video. So, please hit the subscription button, turn on your notification bell, and please comment—this is very important—tell me how you feel about the video you are watching, so that it will help me improve in my next video. Thank you. Alright, so moving forward now, let's do this. We have x plus 1 over x, bracket x squared, minus, now this and this we go, we are left with minus 1, plus 1 over x squared. You know that 1 squared is still 1, and x squared is x squared, and this equals 110. Okay, now let me rewrite this as x plus 1 over x, in parentheses, times x squared, let me bring plus 1 over x squared closer, and move this to the other side. Let's continue. Don't mind me, I continue with serial subtypes. Now, I want to express x squared plus 1 over x squared in terms of x plus 1 over x. Yes, now let's do it this way. Remember that a plus b all squared, right, is equal to a squared plus 2ab plus b squared. Therefore, we can rewrite x plus 1 over x squared as a squared, that is x squared. So it's going to be x plus 1 over x, all squared, plus 2. No, minus 2, because this will come here to become negative. Minus 2. Alright. Moving forward, we now have this in this equation, and we're going to replace this with that. Alright. So I'll write this as x plus 1 over x, open bracket. Now instead of me writing x squared plus 1 over x squared, I'll write this. So I'll now write, I need to clean this, oh my gosh. So I'll now replace this with the entire expression. So I'll now have x plus 1 over x squared, minus 2, then minus 1, equals 110. Let me now clear the rest. Now we can write this as (x + 1/x), enclosed in parentheses, (x + 1/x) squared, minus 3, equals 0. I have achieved my goal. Yes, because I have (x + 1/x) here, and (x + 1/x) here. I can now say, let x plus 1 over x equal y. If I make that statement, then this particular equation becomes y, parentheses, y squared minus 3, close parentheses, equal to 0. Alright. Now let me expand. y times y squared is y to the power of 3, minus 3 times y is minus 3y, is equal to, where am I getting this 0 from? Is equal to 110. So it equals 110. So I'm going to write this down somewhere because it's very important that x plus 1 over x equals y. Alright, now let's move forward. y cubed minus 3y, this comes here to become minus 110, is equal to 0. Now, I want to factorize this. You know how we roll. This minus 110, we'll find a way to put it in cubes. So I have y cubed minus 3y, which will be minus 125 because this is 5 cubed, right? Plus 15 equals 0. So we can now write y cubed minus 3y minus 5 cubed. This 15 in terms of 3 will be minus 3 minus 5, right? Minus 3 times minus 5 equals 0. Now let's bring this together with this. So we now have y cubed minus 5 cubed. Then we have minus 3y minus 3 times minus 5 equals 0. So recall that y cubed minus b cubed, a cubed minus b cubed is equal to a minus b multiplied by a squared plus ab plus b squared. In this case, our a is y, our b is 5. That is for this one. So let's go ahead and write. We are going to write y cubed minus 5 cubed as (y minus 5) times (y squared plus 5y plus 25). That's it for this. Then from here we have minus 3 minus 3. Let's factorize that. So we have minus 3 in parentheses: minus 3y divided by minus 3. We are left with y. And minus 3 minus 5 divided by minus 3. We cancel this; we are left with minus 5. We are almost there. We have y minus 5, y minus 5 here. So we can actually factor that out. So we have y minus 5 times, now this becomes y squared plus 5y plus 5 squared, which is 25. Then minus 3 minus 3. Where on earth am I putting this 0, for heaven's sake. Okay, okay, for now it's 0 because we moved minus 110 to the other side. So it equals 0. So now we have (y minus 5) multiplied by (y squared plus 5y plus 22) equals 0. So from here we have y minus 5. Remember, we are not looking for y, we are actually looking for x. So stay tuned and see. So we have y minus 5 equals 0 and we have y squared plus 5y plus 22 equals 0. So from here, our y is 5, and from here, our y will be minus b, which is minus 5, plus or minus the square root of b squared, which is 5 squared, minus 4a. So a is 1, right? 22ac. Everything divided by 2, multiplied by a, which is 1. So I'm using the quadratic formula here. I believe you know what I mean. So this is 5 squared, 25, and this place is giving us 88. So we are rejecting that. Since this is negative, we are rejecting it. Now we have only one option: y equals 5. Therefore, y is equal to 5. But recall that x plus 1 over x is equal to y. I wrote it somewhere. So since our y is 5, we can write x plus 1 over x is equal to 5. Now to make this a linear equation, let's multiply through by x. I'm trying to write it more concisely. So, x times x is x squared, 1 over x times x is 1, and 5 times x is 5x. So moving forward, we have x squared. This will come out to be minus 5x plus 1 equals 0. Now we are going to solve this using the quadratic equation formula. Now solving this, we say a is 1, b is negative 5, and c is 1, and so on. Now let's solve. So x will be minus negative 5 plus or minus the square root of negative 5 squared minus 4ac. 4 multiplied by 1 multiplied by 1 again. Everything divided by 2a, which is 2 multiplied by 1. So we have x is 5 plus or minus the square root of minus 5 squared, which is 25. 4 times 1 is 4. Everything divided by 2. So now we have x equals 5 plus or minus the square root of 21 divided by 2. So this is the value of our x. Thank you so much for watching, and I'll see you in my next video. Bye.