This video demonstrates how to solve the equation B³/(3B) = 6 by first simplifying to B³ = 18B, then factoring to B(B² - 18) = 0, and finally applying the difference of squares to find solutions B = 3√2 and B = -3√2, while verifying that B = 0 is extraneous.
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Olympiad Mathematics | Indian | Can You Solve This?Added:
Hi, everyone.
Let's provide a complete solution to this equation here.
B * B * B over B + B + B = 6.
What do we do from here?
Um B * B * B is going to give us B to the power of 3.
Right?
Because in this case, we uh adding the powers. And we have invisible powers to the three of them.
So, we add all the powers.
And then for the denominator B + B + B, we are going to add the coefficients.
And each of them has invisible one as a coefficient. So, we add and we get 3 B.
This is equal to 6.
And at this point, you know, we have this over one. So, we can easily cross multiply.
And we have B to the power of 3 to be equal to 18 B.
We have 18 B. There are ways we can deal with this.
Yes, there are actually ways that we can deal with this. Let's bring this to the left. So, we have B to the power of 3 - 18 B being equal to 0.
And to solve it, we have to factorize very quickly.
And our B will come out. Here, we have B squared remaining.
And here, only 18 remains.
And we equate to 0.
At this point, we apply our zero product rule. So, B is zero or B squared minus 18 is equal to zero.
So, as a matter of fact, from here we can say that B equals zero is a solution.
We'll come back to verify whether it satisfies or not.
Then on the other hand, from here B squared minus 18 is zero, right? But I would love us to use difference of two squares. So, we'll write this as the square root of 18 squared.
Okay, this is to help us get difference of two squares over there.
And we have zero.
Let me explain this better.
Square root of X squared is the same thing as X.
So, if you have 18, we can write it as square root of 18 squared. Remember this will take this out.
So, that is what I did over there.
Now, let's remove this.
And I'm going to use difference of two squares. If you have X squared minus Y squared, this is the same thing as X minus Y multiplied by X plus Y.
Right?
So, our X now is going to be B, so we put B there.
Minus the square root of what? Square root of 18.
Then we have B plus the square root of 18.
And everything is equal to zero.
So, at this point, we will apply difference of zero product rule again.
We're multiplying this and this to get zero. So, it's either this is zero or this is zero. So, let's work on this.
Okay, so b minus square root of 18 is equal to zero or b plus the square root of 18 is equal to zero.
From this part, b will be equal to zero plus root 18, and that will give us root 18.
Or from this part, b will be equal to zero minus root 18, and that is minus root uh 18.
Now, what is 18?
18 is the same as 9 * 2.
Here, we have b again to be minus square root of 9 * 2.
Now, according to one of the rules of um sword, we can split this to get square root of 9 * the square root of 2 or b here will be negative square root of 9 * the square root of 2.
So, that here, b will be square root of 9 is 3 * root 2.
Then here we have um -3 cuz square root of 9 is 3 then * root 2.
So, at this point, we have the complete solution.
Remember, before now, we got b to be zero.
And then, we have b, let's say b1, b2 now.
This is b1, b2.
B2 is 3 root 2 and b3 is -3 root 2. So, these are the three solutions.
But, do you think the three of them can satisfy?
Let's check it out.
Okay, so the first solution is B equals zero.
So, if you put zero now, it means you're having zero times zero times zero over zero plus zero plus zero.
And this can never give six.
Okay, as a matter of fact, B to be equal to zero is not a solution.
The second solution is B equals three roots two.
So, this means that we'll have three three roots two times three roots two times times three roots two.
And then it's over three roots two plus three roots two plus three roots two again.
So, will this give us six? Let's try.
Three times three times three, that is 27.
So, let me write 27 here.
And then roots two times roots two is two.
Then two times roots two, that will be two roots two.
Then for the denominator, three plus three plus three, that is nine.
So, we have nine roots two.
Okay, so roots two can cancel roots two.
Then nine into 27 is three.
Three multiplied by two is six. So, it satisfies.
Let's try the the third solution, which is um negative So, B is equal to negative three roots two.
So, to try this, we have negative three roots two multiplied by negative three roots two multiplied by negative three root two.
And this is going to be over minus three root two. Even though it's supposed to be, you know, addition between them, but the negative will overpower. So, we have this root two and we have this root two as well.
Now, if you multiply negative times negative times negative, that will give us negative. Then, three times three times three is 27.
Then, we pick up this times this.
Root two times root two is two. Two times root two is um two root two.
Just like we had before.
Then, minus three minus three is minus six. Minus six minus three is minus nine. So, we have minus nine root two what? Two. This will still go with this.
Negative will cancel itself. Then, three will go nine will go into 27 three times. Then, three times two is two what?
Six. So, that means that the value of B also satisfies.
Therefore, the solution now is B equals three root two as the first and then B two is equal to minus three root two.
Thank you for watching.
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