Germany | A Nice Olympiad Algebra Problem

Ajouté : 2026-05-12

[00:00:01]Hello everyone, you're welcome to solve this nice math problem which is the cube root of 1332 - x ^ 3. This is equal to the roo<unk> of 122 - x ^ 2. So the question is what is the value of x given that x is an element of integers. Now let's provide a solution from here.

[00:00:26]So the first step to do here we can let we can let u to be equal to the cube root of 1332 subtract x ^ of 3.

[00:00:44]This is equal to square<unk> of 122 subtract x ^ of 2. Now this is to mean that we have u is equal to the cube root of 1332 subtract x ^ of 3. Then we have u is equal to the square root of 122 subtract x ^ of 2. From the first part here let this to the power of 3 on both sides.

[00:01:19]So that now here we have u ^ of 3 this is equal to now here we eliminate the cube root sign so that we have 13 32 subtract x ^ of 3. Now let's take -x ^ 3 on the left hand side. So that now we have u ^ of 3 + x ^ of 3 this is equal to 13 32.

[00:01:50]Let's call this as the first equation that is equation one. From the second part that we have here now we have to eliminate the square root sign here. So this means we raised to the power of two on both sides. So that now here we have u ^ of 2. This is equal to we eliminate the square root sign and this means we have 122 subtract x ^ of 2. Now let's take x ^ 2 on the left hand side. So that we have u ^ 2 + x ^ of 2.

[00:02:28]This is equal to 122.

[00:02:33]Now let's call this as equation two.

[00:02:37]Now from equation 2 which is u ^ 2 + x ^ of 2 this is equal to 122.

[00:02:50]Now from what we have here we have that u + x from the algebraic identity u + x ^ 2 this can be expressed as u ^ of 2 + x ^ of 2 then + 2 u x now we have that u ^2 + x 2 this is what we have here 122 so we mean that u + x raised to ^ 2. This is = 122 + 2 ux.

[00:03:28]Now the next step from here we have let's make 2 ux to be the subject of the formula. So this means we have 2 u x this is equal to u + x raised to ^ of 2 then subtract 122 subtract 122. Now the next step is to divide on both sides by two. So that now we have ux is = u + x raised to ^ of 2 this is divided by 2 then subtract 122 / by 2. So this is to mean that u * x this is equal to we have u + x raised to ^ of 2 / 2 then subtract 122 / 2. This is equal to 61. Let's call this as equation equation 3. Now the next step is start from equation one which is u ^ 3 + x ^ 3 this is equal to 1332.

[00:04:49]Now we have that u ^ 3 + x ^ 3. This is actually the sum of two cubes expressed as a ^ 3 + b ^ 3 which we can express as a + b multiplying by a ^ of 2 subtract a b then plus b ^ of 2. Applying this identity, this means that here we have u + x multiplying by u ^ of 2 subtract ux then plus x ^ of 2. This is equal to 13 32.

[00:05:33]The next step is that we have u + x into the parenthesis. This is u ^ of 2 + now + x ^ of 2 - ux.

[00:05:49]This is = 13 32.

[00:05:53]Now we have u + x.

[00:05:57]Here we have u 2 + x 2. Remember u + x2 this is what we have in equation 2 this is what we have in equation 2 here u ^2 + x² is this is 122 so let's substitute here so we have 122 then subtract ux this is equal to 13 32 now the next step is that we have ux is equal to equation 3 which is u + x ^ 2 / 2 - 61. So let's substitute a. So we have u + x * 122 subtract ux which is actually equal to u + x. This is raised to the power of 2. Now this is divided by divided by 2 subtract 61 close the parenthesis. This is equal to 13 32.

[00:07:05]The next step is that we have u + x into the parenthesis. This is 122. Let open the parenthesis here. - u + x raised to the power of 2 / 2 * 61. This is + 61 and this is equal to 13 32.

[00:07:27]The next step is that we have u + x into the parenthesis 122 + 61 this is 180 3 then subtract u + x raised to ^ of 2 / 2. This is equal to 13 32.

[00:07:52]Now the next step is that we have u + x multiplying by now 183 is a whole number. So this is over one. The LCM here is 2. 2 / 1 is 2 * 183. This is 366 subtract x + that is u + x raised to ^ 2 this is equal to 13 32.

[00:08:20]So we have u + x multiplying by this is 366 subtract u + x ^ of 2. This is equal to 13 20 132.

[00:08:37]Now let's let's eliminate these two.

[00:08:40]Let's multiply both sides by 2. So that now here we have u + x.

[00:08:49]This is multip 366 subtract u + x. This is raised ^ of 2.

[00:09:00]Then this is equal to 13.2 * 2. This is 26 64.

[00:09:08]Now we have that u + x is common here.

[00:09:11]So we can let u + x be = k. So let's let's express this equation in terms of K. So we have K * 366 subtract K ^ of 2. This is equal to 26 64.

[00:09:33]Now the next step is that K * 366 this is 366 Kus K ^ of 3. This is equal to 264.

[00:09:43]Let's take 2664 on the left hand side.

[00:09:46]So that we have 366 k minus k ^ of 3 subtract 26 64 this is equal to zero. Let's rearrange this equation so that we have k ^ 3 + 366 k subtract 26 64 this is equal to zero.

[00:10:09]Now let's multiply everything here by1.

[00:10:13]So we have -1 * k ^ 3 this is k ^ 3 + 366 k * -1 this is - 366 k then + 26 64 this is equal to 0 *1 this is equal to 0. So we form a cubic equation here. We form a polomial here of degree 3. This is a cubic equation.

[00:10:39]Now we have a cubic equation here. Now let's apply the rational root theorem where we have k is equal to now we have + or minus1 + or minus 2 all the way to + or minus 12. Now given that k is = 12.

[00:10:58]Let's substitute 12 in this cub equation. And this means we have 12 ^ 3 - 3 66 multiplying by 12 + 26 64.

[00:11:13]This should give us a value of 0. So we have that 12 ^ 3 this is 17 28 - 366 * 12 which is equal to this is the same thing as 4392 then + 26 64 this is equal to zero. So we have 1728 + 2664 this is 4392.

[00:11:44]Let's subtract 43 92 and this should give us a value zero. So we have that 0 is equal to zero and this means that k = 12 this is one of the roots. Now from the cubic equation which is k ^ 3 - 366 k + 2664 this is equal to zero. We can express this equation as k ^ 3. Now let's introduce -2 k ^ of 2 + 12 k ^ of 2 subtract 366 k then plus 2664 this is equal to zero. Now from the first part here we have that k to the^ of 2 is common. So let's factor out k^ 2. So that we have k ^ 3 / k² this is k - 12 k² / k 2 this is - 12 then plus now here we have 12 k^ 2 we can express - 366 k this is - 144 k then subtract 222 k then + 2664.

[00:13:19]This is equal to zero.

[00:13:22]We have -44k - 222 K. This is the same thing as - 366 K. Okay. The next step is that we have K ^ of 2 K subtract 12 then plus in the second part here we have that 12 12k is common. Let's factor out 12k. 12k ^ 2 / 12k this is k subtract 12 then minus in the third part here we have 222 it's common here. So let's factor out 222.

[00:14:02]So that now here we have k subtract 12. This is equal to zero. So you can see clearly that k minus 12 is common here. So let's factor out k - 12 into the parenthesis. This is k² plus here we have 12k then we have subtract 222.

[00:14:31]This is equal to zero.

[00:14:36]Now we have got two parts here. we have k - 12 this is equal to z then we have k ^ 2 + 12 k - 222 this is equal to to zero from the first part k is = 12 now we have a quadratic equation here where we have a is = 1 b is = 12 and we have c is = -222 Now the next step is to assess the nature of the root of this quadratic equation. And that means let's assess the discriminant value which is b ^ 2 - 4 c. So let's substitute a, b and c in this formula here. So we have 12^ 2 - 4 * a which is 1 * c which is 222.

[00:15:31]So what we have is that the discriminant is= 12 which is 144.

[00:15:37]then plus 4 * 222 this is 888 and what this means is that we have the discriminant value which is equal to 1032 which you can also express as 129 * 8.

[00:15:58]Now we have the discriminant value is non negative number. But now this is not a perfect square. This is not a perfect square.

[00:16:11]This is not a perfect square.

[00:16:14]And therefore and therefore this is a non integer solution.

[00:16:20]This is a non integer solution.

[00:16:24]Since we have that x comma u and k these are members of integers.

[00:16:34]Now given that this is not a perfect square. So this means that the quadratic equation here is rejected. This quadratic equation is rejected.

[00:16:48]Now let's focus on k = 12. k = 12. Now if you recall, if you recall, we understand that we let that is u + x be = k and in this case k is = 12. Now from equation three, from equation three, equation 3 is equal to ux which is = u + x raised ^ of 2 / 2 - 61. This is equation three as we noted earlier.

[00:17:33]So now we can substitute u + x which is 12 in this equation. So that we solve for ux.

[00:17:41]Now let's substitute u + x so that ux is = 12 raised ^ 2 / 2 - 61. So that now ux is = 12² this is 144 / 2 - 61.

[00:18:00]So we have that ux is = 144 / 2 this is 72 - 61 and this is equal to 11. So we have that ux is equal to 11.

[00:18:16]Now we have that u + x this is equal to 12.

[00:18:24]Now from here let's make u to be the subject of the formula. So that means that u is = 11 / x. So we divide both sides by x. So we have u is = 11 /x. So let's substitute u which is 11 / x here.

[00:18:42]So we have 11 / x + x. This is equal to 12. Now x is our number and 12 is our number. So let's multiply hcm here which is x. Let's multiply everything by x. So that now we have 11 + x * x. This is x ^ 2 and this is = 12 x. Let's take 12 x on the left hand side. So that we have x ^ of 2 - 12x + 11. This is equal to zero. So this is a quadratic equation. We can solve this quadratic equation by factorization method. we have x ^ 2 - 11 x - x then we have + 11 this is equal to to zero we have got two parts here from the first part x is common let's factor out x so that we have x - 11 then subtract now here one is common let's factor out one so that we have x subtract 11 this is equal to to zero so we I have got x - 11 is common. Let's factor out x - 11. And then we have x -1. This is = 0. Okay.

[00:20:02]Now we have x - 11 * x - 1 is equal to 0. So this is x - 11 = 0. Then we have x -1. This is = to 0. So we have got two possible values of x. x1 is = 11.

[00:20:19]and we have x2 this is equal to 1. So these are the two values of x for this algebra problem.

[00:20:29]Now let's check if these two values of x satisfies the equation. Now let's start with x1 which is 11. Now if you recall, if you recall, we have that the cube root of 13 32 subtract x ^ of 3. This should be equal to the square root of 122 subtract x ^ of 2. Let's start with x1 which is 11 in this case. So we have the cube root of 13 32 - 11 ^ 3. This should be equal to the square<unk> of 122 - 11^ 2.

[00:21:24]Now we have the cube root of 1332 minus 11 ^ 3. This is 1331.

[00:21:36]This should be equal to the square root of 122 subtract 11 squar which is 121.

[00:21:46]So we have the cube root of 13.2 - 13 that1 this is the cube root of 1. This should be equal to the square root of 122 - 121. This is the square root of 1.

[00:22:00]The cube root of 1. This is one which is equal to square root of 1 which is 1. So this means that the left add side is equal to the right add side and this affirms that x1 which is 11 satisfies the equation. Let's check x2 which is equal to one from here. Now let's check x2 which is equal to 1. If this satisfies the equation, now we'll have the cube root of here we have 13 32 - 1 ^ 3. This should be equal to the square root of 122 subtract 1 ^ of 2.

[00:22:49]Now this is the cube root of 13 32 subtract 1 ^ 3. This is 1. This should be equal to square<unk> of 122 subtract 1 2 which is 1.

[00:23:08]So we have this is the cube root of 13 32 - 1 this is 13 31. This should be equal to the the square root of 122 - 1. This is the square root of 121. Now we have the cube root of 1331. This is the same thing as 11 ^ 3. This should be equal to the square root of 121 which is 11 ^ of 2. Now if we eliminate the cube root sign here we have 11 is equal to now again we eliminate the square root sign 11 is equal to 11 and this indicates that the left add side is equal to the right add side and this affirms that x2 which is equal to 1 also satisfies the equation. So we have got two solutions x1 which is 11 x2 which is 1 for this algebra problem. So kindly follow the steps like this video and subscribe.