Solve for x in this nice Algebra equation | Math Olympiad Mathematics

Added: 2026-05-05

[00:00:01]In this video, let us find the value of x given x raised to power 6 is equal to 4 raised to power 6.

[00:00:22]We have x raised to power 6 is = 4 raised to power 6.

[00:00:29]The first thing to do will be to move everything to the left hand side. So we have x^ 6 - 4 raised to power 6 is = 0.

[00:00:43]Then knowing that 6 is equivalent to 3 * 2, we'll replace 6 with 3 * 2 on both terms. So this is x raised to power 3 * 2 - 4 raised to power 3 * 2 is = 0.

[00:01:10]A raised power m * n by law of indices can also be written as a raised to power m then raised to power n then we can express this as x raised to power 3 raised power 2 - 4 raised power 3 raised power 2 = z.

[00:01:39]We see difference of two squares given a raised to power 2 minus b raised to power 2. This is difference of two squares and by identity this can be factorized as a minus b into bracket a + b.

[00:02:03]So all of this becomes x raised to power 3 - 4 ra^ 3 into brackets x raised to power 3 + 4^ 3 is equal to z.

[00:02:26]This will imply that either x^ 3 - 4^ 3 is = 0 or x raised to power 3 + 4^ 3 is equal to z.

[00:02:45]These are two different situations. For the first one, we have difference of two cubes and here we have addition of two cubes. Now let us solve for x in each of this separately. So we'll call this case one and call this case two.

[00:03:13]So starting with case one we have difference of two cubes and by identity given a raised to power 3 - b raised to power 3. We can factoriize this as a minus b time a 2 + a * b + b ^ 2.

[00:03:43]So this equation becomes x - 4 into brackets x^2 + 4 x + 4 2 = 0.

[00:04:03]x - 4 into bracket x^2 + 4x + 4² here is 16 = 0.

[00:04:20]This will imply that either x - 4 is = 0 or x^2 + 4x + 16 is = 0.

[00:04:40]From here x is going to give us 4.

[00:04:45]This is our first solution from this problem. x1 = 4 and this is real solution.

[00:04:57]Then we go to this quadratic equation to solve for x here. We compare it first with a x^2 + b x + c = 0.

[00:05:11]A is going to be 1, B is going to be four and C is going to be 16.

[00:05:21]Then we use the quadratic formula X = - B plus or minus B ^ 2 - 4 A C / 2 A.

[00:05:40]Let us plug all these values into this formula.

[00:05:45]This will give us x is = minus b is 4.

[00:05:52]So we have -4 + minus square<unk> of 4^ 2 - 4. We know that a is 1 and c is 16.

[00:06:12]2 a that's 2 * 1 then x is = -4 + - square root of this will be 16 minus this will be 4 * 16 / So we have x is = -4 plus or minus the square root of I'm going to factoriize 16 out into this becomes 1 - 4 then / This becomes x is = -4 plus or minus square<unk> of 16 * -3 / 2.

[00:07:24]Then x becomes -4 plus or minus. We separate this radical into square<unk> of 16 *<unk> -3 then / 2 giving us x is = -4 + roo<unk> of 16 here is 4. So we have here also we can split this into square<unk> of 3 *<unk> of -1 then / 2 x is = -4 + or minus root of -1 is i. So we're going to have 4 I <unk>3 4 I <unk>3 then / 2.

[00:08:24]Now when we separate this division this will give us X is = - 4 / 2 + or minus 4 I <unk>3 / 2.

[00:08:40]So two here is one, two here is two, two here is one, two here is two. So x is = -2 + or minus 2 i <unk>3 giving us x2 is = - 2 + 2 i <unk>3 and x3 - 2 - 2 I <unk>e 3 which are both conjugates.

[00:09:26]These are complex solutions.

[00:09:33]Now remember we got a real solution earlier which was x1 and the value was four and this is real.

[00:09:45]We call this real solution or real root.

[00:09:50]So we've gotten three values from the first case where we have difference of two cubes. Now let's see what we'll get from the second case which is addition of two cubes. Similar to how we solved the first case, this time we have addition. So given a raised to power 3 + b raised to power 3 by identity we can factoriize this as a + b into brackets a 2 - a + b 2.

[00:10:33]Then this equation becomes x + 4 into bracket x^2 - 4x + 4^ 2 = 0 x + 4 into bracket x^2 - 4x + 4^ 2 here is 16 = 0.

[00:11:11]This will imply either x + 4 is = 0 or x^2 - 4x + 16 is = 0.

[00:11:28]From here we have x is = -4.

[00:11:33]This will be x4 cuz we've gotten three other values for x earlier.

[00:11:41]Let us solve for x from this quadratic equation. Here a is = 1, b = -4 and c is 16.

[00:11:57]So x is going to be minus b. B in this case is -4 then plus or minus square<unk> of - 4^ 2 - 4 * A is 1 and C is 16.

[00:12:24]all of that divided by two * 1.

[00:12:31]So we have x is equal to minus * minus here is 4 is positive. So we have pos4 plus or minus the square root of this is 16 - 4 * 16 / 2 x = 4 plus or minus square root of again we factor 16 out into 1 - 4 square root of that then / 2.

[00:13:21]This is x = 4 + -<unk> of 16 * 1 - 4 is - 3 then / 2.

[00:13:38]So x is = 4 + minus again we separate this into square<unk> of this will be 16 * - 3 is 3 * -1 so we have<unk> 3 *<unk> -1 then / 2 so x is = 4 plus or minus this will be four.

[00:14:12]We said this is I then this is <unk>3 then divided by two.

[00:14:22]Then we separate the division to give us x is = 4 / 2 + or - 4 i <unk>3 / 2 here is 1 2 here is two two here one two here two so x is = 2 + - 2 i roo<unk> 3 and then we can separate this into x5 = 2 + 2 i <unk>3 and x6 2 - 2 i roo<unk>3.

[00:15:18]So we got two more complex solutions and one additional root which is x4 which we got -4.

[00:15:37]So this is real solution. So together we got two real solutions and four complex roots from this problem. Bringing us to the end of the video. If you have enjoyed this video, please give it a thumbs up and remember to share and subscribe to my channel if you have not done so already.

[00:16:00]And I'll see you in my next video. Bye.