Olympiad Mathematics | Japanese | Can you solve this?

Añadido: 2026-05-11

[00:00:00]If you're ready, let's provide the solution to this one.

[00:00:08]M to the^ 3 + m to the^ of 2 = 1 / 8.

[00:00:17]Okay, we want to provide the complete solution to this problem. And our challenge for now is 1 / 8.

[00:00:27]The first step to take is to clear out the fraction. And to clear out the fraction, the LCM is what? 8. So we're going to multiply all through by the LCM. We have our 8 * m then plus the same 8 * m 2 and it's equal to 8 * 1 / 8.

[00:00:53]So from here now we can cancel out eight on the right part. So we have only one there. But 8 can be written as 2 ^ 3. So this is multiplying m ^ 3 plus look at this 8 is the same as 2 * 4 then multiply by m² and everything is equal to 1 like I rightly said but then here we have 2 to the^ 3 * m ^ 3 + 2 * 2 2 4 is 2 Now I'm trying to ensure that the highest power here is 2 because we have m 2. Okay. So we multiply by m² and this is equal to 1.

[00:01:50]Now we are going to apply some laws of logarithm. Remember there's a law of logarithm that says if you have a to the power of m okay there's m in the equation a to the power of b multiply by let's say you have c to the power of the same b. Now there's a law that says you can multiply a by c and then raise both of them to the power of b. Right? This is from one of the laws of indices. So I'm going to apply this same law and you know solve this problem. So watch me apply the law from here. Now I'll multiply 2 and m to get to m and both of them will be raised to the power of what? 3.

[00:02:44]Then plus here I will have 2. Then 2^ 2 * m 2 I'll multiply both of them. I have 2 m both of them to the power of two and everything is giving us one on the other side of the equation.

[00:03:05]Okay. So from here now you will see that we already have a common term and that is 2 m you know in brackets 2 m 2 m over there. So we can say that let y be = 2 m. So that if y is 2 m we will be having y cub plus here we have 2 y^ 2 and let's bring this to the left so it becomes minus1. Now everything is zero.

[00:03:40]Now how do we solve this problem here?

[00:03:43]there's no common you know no common factor right nothing is common to the three of them so what should we do let's split these two so that we will practice what we call group factorization so I'm going to write y cub plus to split this you're going to have y^ 2 + y^ 2 then -1 = 0 now the group factorization is here right I will factoriize these two what is common to this two is um y^ 2 then here there's going to be one y and here we are going to have just one right then plus here we have y^ 2 - 1 2 that is difference of two squares so let me express it properly so we're going to have y 2 into y + 1 + from difference of two squares we have y um + 1 into y - one from this part and everything is equal to zero.

[00:05:00]So what again do we do factoriize because we have something common to the two of them and that is y + 1. So y + 1 comes out as the common factor common factor and then here we have y² + remember this is going to be y - one because this one here is already out. So this is equal to zero and we can apply our zero product uh rule because we are multiplying these two to get zero and every time you multiply two terms to get zero either of them will be equal to zero. So let's do that.

[00:05:52]Okay. So we're applying our zero product rule. So y + 1 is 0 or y^ 2 + y - 1 is = 0. And from here y = 0 -1. So y = -1.

[00:06:13]This is one of the solutions.

[00:06:16]But mind you, we did not even have y in the equation. Right? We will use this to get the value of m. Now let us go back and pick this um particular expression and then equation and solve it. Remember that is a quadratic equation and we are going to use this formula for it. y = - we have b^ 2 - 4 a c under the square root sign and it's all over 2 a. From here our a is 1, our b is 1 and our c is min -1.

[00:06:56]a b c right? So we are going to enter these values into this formula and y will now be in place of minus b we write -1 since our b is 1 + minus we have b^ 2 which will be 1^ 2 then - 4 * a what is a a is 1 then multiply by c which is -1 then this is all over 2 * 1.

[00:07:31]So to go on with this, we're going to have y to be equal to -1 + or minus the square root of 1. 1 2 is 1. 4 * - 4 * -1 is + 4.

[00:07:52]And this is all over 2.

[00:07:56]To proceed with this, we will have y to be equal to -1 + or minus square<unk> of 5 over 2. This is a two in one value of y because from here because from here we have that we have our y to be equal to -1 +<unk> 5 / 2 as one of the solutions or y to be equal to -1 -<unk> 5 over two, right? But mind you, we were not looking for the value of y originally. So we have to go back and get our corresponding values of m.

[00:08:51]Okay. So we're going to bring down the three values of y.

[00:08:56]We're going to bring them down here. We got y to be equal to minus1.

[00:09:02]That is one of the values of y. Then we also have um -1 +<unk> 5 / 2. That's the second value of y. Then the third value of y is -1 -<unk> 5 / 2. Now these three values of y are going to have their corresponding values of m because we say that let y be equal to 2 m. So now when y is 2 m here I'm going to write 2 m to be = -1.

[00:09:43]Then divide both sides by 2ide by two.

[00:09:47]Two will go. So our m is um equal to -1 / 2. So at this point we have our first value of m which we call the m1.

[00:10:02]Right? Now let's go get the other two values of m.

[00:10:08]Okay. So remember that we still have y to be that. Right? So to bring it down, we're going to write our 2 m, which is the value of y now to be equal to -1 +<unk> 5 / 2. So what do we do here? We divide both sides by two. This will take this out. So our m by the way, we still divide this one by two, right? So our m from here will be equal to -1 +<unk> 5 over 2 * 2. This two we multiply this two and we have four. This is another value of um m. We will call this our m2. Now the only difference between this value and the third solution is that this place is going to be negative in the third solution. So our m3 will just be -1 -<unk> 5 / 2. So by this we have solved the equation completely and I believe you enjoyed yourself. If you did, subscribe to my channel for more.