The video provides a lucid and methodical breakdown of complex root derivation through disciplined algebraic factoring. It is a precise demonstration of how structural logic can simplify and solve seemingly intimidating equations.
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Olympiad Mathematics | How I got the four solutionsAdded:
I hope you know how to solve this problem here.
Okay, if you do not know, then let's solve this together.
m to the power 4 equals um 4 m to the power of 1 over 3 everything to the power of 3.
Now, the first thing that will come to the mind um the minds of students is that they will um cancel this 3 and this 3 and that will be completely wrong.
Yes. Because what we have here, remember that if you have a b to the power of c, this is expressed as a to power c times b to power c.
Meaning that both 4 and m here are having to do with this power 3.
So, see how you should simplify that.
Okay, so you should have m to the power of 4 to be equal to 4 to the power of 3 times m to power 1 over 3 then to the power of 3 as well.
Right?
So, from here now we have our m to the power of 4 and it's equal to 4 to the power of 3 then this 3 can go with this 3 and we just multiply by by m.
Now, [snorts] do not divide both sides by m. What do you do? Bring this to the left-hand side so that we can have m to power 4 minus 4 to power 3 m all equal to 0.
So, what do we do?
Remember, how many solutions do you think we are going to have?
Is it three or four?
We are going to have four solutions.
Why? Because the highest power is four.
So, we're getting four solutions. Now, the first thing you're going to do from here is to factor out your M.
So, M is a common factor. Here, we have M to the power three remaining.
>> [snorts] >> Minus here, we have four to the power of three. Remember that this is all equal to zero.
Yes.
And [snorts] at this point, we apply our zero product rule to say that M is equal to zero or M to the power of three minus four to the power of three is equal to zero.
If this is true, then we have a solution already, which is M equals three.
And like I told you, we are looking for three solutions. Four solutions, right?
So, we're going to get three more solutions from here.
What do we do?
We write our um Let me write it here. M to the power of three minus four to the power of three. This is um equal to zero.
And it is a difference of two cubes.
Mind you, A to power three minus B to the power of three is still the difference of two cubes. And it's A minus B into A squared plus AB plus B squared.
This is our difference of two cubes identity.
So, our A now is going to be M and our B is going to be four. So, in place of A minus B, we are going to write M minus four.
And in place of A squared, we are writing M squared.
AB, that's going to be four * m, and it's 4 m.
Then we have + b squared. Our b squared is what?
Our b squared is going to be 4 squared, which is 16.
Right?
16. By the way, there's something I would like to do. Instead of writing 16, let me write 4 squared as well.
Okay, b squared is going to be 4 squared.
And this is equal to 0.
Now, I have my reasons for writing 4 squared instead of 16.
What do we do?
We apply our zero product rule.
So, that m - 4 is equal to 0.
Or m squared + 4 m + 4 squared is equal to 0.
So, we will come back to this.
But, before then, from here, we have our m to be 0 + 4.
And the value of m is equal to 4. We have our second solution.
Yeah, we're going to bring the solutions together after.
Now, here we have a quadratic equation, which will give us two more solutions.
What do I do?
Let's bring it out, and then we solve it.
Okay, so this is it.
The quadratic equation, and we will solve this using the quadratic formula.
Now, our a is 1, coefficient of m squared.
And um our b is 4, the coefficient of m.
And the value of c is 4 squared.
Okay, I want to use it as 4 squared.
Now, what is the formula?
m is equal to - b + - we have b squared - 4 ac This is all over two times A.
So, what we'll do now is to put in the values of ABC.
So, that our M will be In place of minus B, we'll put minus four.
Plus or minus B squared is here as four squared.
Then minus we have four times one.
Right? We have four times one. Then we multiply by four squared.
Because C is four squared from there.
This is all over two multiplied by one.
And it will give us one um it will give us two eventually.
>> [snorts] >> From here, we're going to get M to be minus four plus or minus If you look at this Okay, so from here, we have um four squared.
And four squared as a common factor. So, four squared will come out.
And that was the reason I left this as um four squared.
Then we open brackets.
Four squared divided by itself is one.
Four times one is four.
And this four squared is already out.
So, we divide this by two times one, which is two.
So, from here, we are getting our M to be equal to minus four plus or minus we have the square root of four squared multiplied by minus three.
One minus four.
And we have this over two.
Now, you can split what you have under the square root sign.
So, [snorts] that M will be minus four plus or minus we have the square root of 4 squared multiplied by the square root of -1 um 3.
And it's all over what? 2.
So that if we go on from here, we're going to get m to be equal to -4 plus or minus square root of this will go.
Okay. The square root and the square will go, so we have from 4 then multiply by this negative here is going to come out and we write the square root of -1 as i.
So what we'll have then now is root 3.
The negative is no longer there and we are dividing by 2 just like this.
From here, our m will be -4 plus or minus 2 into Oh, let us divide this already.
2 into -4 is -2 plus or minus 2 into 4i, that's going to be 2i then we have root 3.
By the way, this is a two-in-one solution.
Like I said, we'll bring the four solutions together.
Before now we got our m1 to be zero.
Okay, let's write that.
m1 is zero then our m2 we got it to be four.
Our second solution.
Then our third solution is equal to from here we have -2 plus 2i then root 3.
Then the fourth solution m4 is equal to -2.
Can you see it?
Okay, -2 -2i root 3.
So these are the four solutions to the given equation.
Thank you for watching.
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