Solving a 'Harvard' University Entrance Exam Question | x=?

Added: 2026-05-12

[00:00:00]Hello everyone, you are welcome. Today we have a very interesting math question. Here is x * x * x - x * x = 18. I will try to find out the value of x and all possible solutions of this interesting algebra math problem. So let's start our solution. First of all to the left hand side here this x is 3 * having power one. So therefore here this expression in the left hand side this will become this is simply x cq minus here this x is 2 * this will become x² and here we will take this expression to the left hand side so it will become 18 is equal to 0. Next we can write this 18 as this is simply this is x cq - x² - this 18 can be written as 27 - 9 is equal to 0 and multiply this negative sign inside the parenthesis. This will become this is xq - x² and this will become - 27 and this will become positive 9 is equal to zero.

[00:01:17]So next we can write these two constant numbers as this is x cq - x² and this 27 can be written as this is simply 3 cube + 9 can be written as 3² is equal to 0.

[00:01:37]Just combine the like terms the same terms the cubic terms. So this will become this will become x cq - 3 cq - x² + 3² is equal to zero.

[00:01:57]So further we can write this expression in the left hand side as this is x cq - 3 cq and from here we will take the negative sign common. So this become x² - 3² is equal to zero.

[00:02:17]And at the left hand side we will use two different algebraic identities. Here in this one expression we will use this one algebraic identity aquus bq that is a - b * a² + a b + b² and here in this one expression we will use this one algebraic identity a² - b² that is a - b * a + b. So using these identities here this expression will become here the first expression will become this will be x - 3 * x² + 3x + 3² minus and using this identity here this expression will become this become x - 3 * x + 3 is equal to zero.

[00:03:27]Look at to both expression both the terms here this x - 3 is common. So we'll take out common. So taking x - 3 common here this will become this will become this is x² + 3 x + 3² simply 9us here only this expression is left x + 3 is equal to 0.

[00:03:56]multiply this negative sign inside the parenthesis. So this will become this is simply x - 3 * x² + 3 x + 9 and this will become - x - 3 is equal to 0. So further let's simplify this expression. So this is x - 3 * x² 3x - 6 it is simply 2 x and here 9 - 3 it is simply + 6 is equal to 0. Here the product of these two expression is zero.

[00:04:38]So here either this linear expression will be zero or this one quadratic expression will be zero. From here we'll get x - 3 =0 or x² + 2 x + 6 is equal to 0. So first we will solve this one equation here. We will take this three to the right hand side. So this will become x = positive3. And this is our first real solution.

[00:05:07]Now we'll try to solve this one quadratic equation. Here we cannot solve this quadratic equation by factorization method. So therefore here we will use the quadratic formula. So for that here we need the values of a, b and c. So here the value of a is simply one. The value of b is 2 and the value of c is 6.

[00:05:28]Now the quadratic formula is x = this is -b + - roo<unk> of b² - power a c / 2 a.

[00:05:41]Now let's substitute the values from here. So this will become the value of x will become this isative b is 2 + minus square root of here b is 2. So 2 square distinctly this is 4 minus this is 4 * here a is 1 and c is 6 divided by 2 * a here the value of a is 1. So 2 * 1 is it is 2. So further this right hand side will become this is -2 + minus square root of there is power common. So we'll take out power common. This become 1 - here 1 * 6 is 6 1 - 6 it is simply - 5 divided by 2. So further this will become x = this is -2 + minus this will become square root of 4 * roo<unk> of -5 can be written as roo<unk> of 5 * -1 divided by 2. So further this expression will become this is simply -2 + minus here roo<unk> of 4 it is simply 2 and this will become square roo<unk> of 5 *<unk> of -1 square root of -1 it is simply that is iota i divided by 2 and there is two common in the numerator so we'll take out two common so by taking two common from the numerator this will become -1 plus -<unk> 5 I / 2 and we'll cancel 2 with 2. So the final values of x will become that is x = -1 + -<unk> 5 I. So here we have two possible values of a and one value is there. So finally the final three values of x are let us suppose x1 is = 3 that is the real value and x2 will become that is -1 +<unk> 5 i and the third solution become x3 is equal to that is -1<unk> 5 i. So finally here we have these three possible solutions and these three possible values of x in this interesting math and algebra math cluster.