Olympiad Mathematics | Indian Can You Solve This One?

Added: 2026-06-03

[00:00:01]Hi everyone. Let's solve this completely.

[00:00:08]X plus one to the power of three equals 64.

[00:00:14]Okay, our target is to look at 64 and see if it is if it is a perfect cube.

[00:00:24]But 64 is a perfect cube, right?

[00:00:27]Yes, because it is four times four times four.

[00:00:32]So, it is a perfect cube.

[00:00:35]And now X plus one to the power of three is equal to four to the power of three.

[00:00:43]So, we have the same powers already.

[00:00:46]But we don't want to equate the bases alone because if you do that, we're going to have just one solution. So, we need to get three solutions to this equation here.

[00:00:59]So, you will say that X plus one to the power of three minus four to the power of three is equal to zero.

[00:01:11]Now, we will now come down to we will come down to what we have and express it in this form.

[00:01:21]You know, if you have a cube minus b cube we can write this as a minus b into a squared plus a b plus b squared.

[00:01:37]You know, we can write this like this.

[00:01:40]So, in that case our a now is going to be x plus one and b is equal to four.

[00:01:51]So, what we will do now is to write put these values of a and b into the identity.

[00:01:58]So, a minus b now becomes x plus one.

[00:02:03]Then, this is a alone, then b is four. So, we put minus four.

[00:02:09]This will be in the first bracket.

[00:02:11]Then, in the second, we have a squared, which will be x plus one.

[00:02:16]And there's a square on this.

[00:02:19]Then, we have plus ab, that will be x plus one for a multiplied by b, which is four.

[00:02:28]Then, we have plus four squared. Four squared is 16.

[00:02:32]Let me just put 16 there, right?

[00:02:35]So, our four squared is 16.

[00:02:38]And everything here, remember, is equal to zero.

[00:02:42]Yes, everything there is equal to zero.

[00:02:45]Now, from here, x plus one minus four is the same thing as x minus three.

[00:02:54]So, this is one of the factors now.

[00:02:57]Then, to get the other factors, you know, we have to expand this one.

[00:03:03]And the expansion of this will give us x squared plus two x plus one.

[00:03:13]That is the expansion of this. Then, plus we need to expand this two. x times four is four x.

[00:03:22]One times four and that is four.

[00:03:26]Then, we have plus 16.

[00:03:30]We equate.

[00:03:31]You know, we're going to equate this to zero. Let me write it here.

[00:03:35]So, everything will be equal to zero.

[00:03:38]Now, let's simplify what we have.

[00:03:42]x minus three is still coming out as the factor.

[00:03:47]Then, here we have x squared.

[00:03:51]Then, two x plus 4 x that will be 6 x.

[00:03:57]Then, we have 1 + 4 + um Okay, let's start from here. 16 + 4 20 20 + 1 that is 21. So, here we have + 21.

[00:04:12]So, we equate to zero.

[00:04:16]Let's go over that again.

[00:04:17]x squared is out. 2 x + 4 x is 6 x.

[00:04:21]Then, 1 + 4 is 5. 5 + 16 is 21.

[00:04:27]This is good. Now, we apply our zero product rule.

[00:04:31]Either this is equal to zero or this one is equal to zero.

[00:04:36]Working with the first factor our x - 3 is equal to zero.

[00:04:43]And this means that x is equal to 3. So, this becomes our first solution.

[00:04:52]This is our first solution. And to get the other solutions, we're going to bring down this um We'll bring down this expression and equate to zero.

[00:05:04]So, that it becomes a quadratic equation.

[00:05:09]Okay, so we We have equated this to zero.

[00:05:15]And now, let's solve it, right?

[00:05:17]Using the quadratic formula so that our a is 1 our b is 6 and our c is 21.

[00:05:28]What is the quadratic formula we're going to use? It is x equals - b + minus square root of b squared - 4 ac all over 2 * a.

[00:05:42]This is the quadratic um formula.

[00:05:45]So, that our x will be in place of minus B, we put our minus six plus minus.

[00:05:53]We have C squared right B squared right and that's going to be six squared.

[00:05:58]Minus four times one times 21.

[00:06:04]Because C is 21. Then we divide all of this by two times one which will give two.

[00:06:11]Now we go ahead to get X to be minus six plus minus.

[00:06:18]Six squared is 36.

[00:06:20]Minus four times one times 21 is 81.

[00:06:26]Right? So this is over two.

[00:06:28]And this is the mistake that some students will make knowing that these two are perfect squares. They'll take the square root of both of them separately.

[00:06:38]And that will be wrong.

[00:06:40]Okay? That will be very very wrong.

[00:06:44]Okay? So see what we should do.

[00:06:47]X is equal to We have minus six plus minus the square root of 36 minus um 81 is minus 45.

[00:06:59]This is over two.

[00:07:03]And now X is equal to minus six plus or minus We have the square root of 45.

[00:07:12]Square root of 45.

[00:07:15]Um yes.

[00:07:17]Minus six plus or minus square root of 45 then multiply by the square root of negative one.

[00:07:24]What I've done is to bring out the square root of the negative number there.

[00:07:29]Brought out the negative here as this then 45 is alone.

[00:07:34]And all of this is over two.

[00:07:38]So our X will now we'll minus six plus or minus 45 is the same as 9 * 9 * 5.

[00:07:47]>> [snorts] >> So, we write this as 9 * 5.

[00:07:52]Then, multiply by square root of -1, which is I.

[00:07:57]Then, we divide by divide by two, right?

[00:08:05]So, to continue with this, we have X to be equal to We have - 6 plus or minus square root of 9 is 3.

[00:08:16]Multiply by I, we have 3i. Then, we have our root 5.

[00:08:21]So, all of this All of this will be over two.

[00:08:27]Okay? So, you can decide to stop here, or let's to divide six and then you manipulate.

[00:08:34]But, let's stop here and bring the three solutions together.

[00:08:38]Okay?

[00:08:41]Okay. So, this is the equation that we have solved.

[00:08:44]And we have our X to be three, the first solution.

[00:08:50]Then, the others, we have X to be equal to - 6 plus 3i root 5 over three.

[00:09:02]It's over two, by the way.

[00:09:05]I've fixed the positive, so the second the next one will be negative now.

[00:09:10]So, we have X3 to be - 6 - 3i root 5 and it's over two. So, these three are the solutions to the given equation.

[00:09:26]Thank you for watching.