To solve equations like Y^3/2 = 4, first eliminate the denominator by cross-multiplying to get Y^3 = 8, then rearrange to Y^3 - 8 = 0, and factor using the difference of cubes formula (A^3 - B^3 = (A-B)(A^2 + AB + B^2)) to find the real solution Y = 2 and complex solutions Y = -1 ± i√3.
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Germany | Maths Olympiad | Can You Solve this?Added:
Hello everybody, welcome to this platform and today's video we have given a question.
Y ^ 3 over 2 equal to 4. We are supposed to find here three values of Y.
Let's provide solution very quickly to this question provided right here. First thing here, we know that we have to copy Y ^ 3 over 2 equal to 4.
And of course here we have down here the denominator of 1 so that we can make cross multiplication to get >> [clears throat] >> Y ^ 3 equal to 2 * 4 equals to 8.
>> [clears throat] >> By grouping like terms here, we shall write Y ^ 3 minus 8 equal to 0.
Now here, what can we do?
We have Y ^ 3 and this 8 can be changed into uh base 2 to get -2 ^ 3 equal to 0. Because 2 ^ 3 equals to 8.
Now, Y ^ 3 - 2 ^ 3 uh is the same as writing A ^ 3 minus B ^ 3.
And of course this one is equal to A minus B times here A squared plus A B plus B squared.
And this is what we are going to to apply right here to get what?
Y minus 2 times what we have right here is equal to Y squared.
Here it'd be plus 2 Y as AB.
Then here plus four equal to zero.
Then, of course, we shall get Y minus two equal to zero or Y squared plus two Y plus four will be zero.
And of course here, Y will be equal to two.
And here, of course, we have a quadratic equation where we shall write Y squared plus two Y plus four will be zero.
So, that here I have to find this is Y1 Y2 and three will be negative B plus or minus here times square root of here B squared minus four times A times C.
Then over two A.
Then, this will be Y squared Y2 and three will be negative two plus minus square root of here will be four minus four times four over two.
And Y2 and three here will be negative two plus or minus square root of here this four minus 16 negative 12 over over two.
And here have Y2 and three will be negative two plus or minus here this negative 12 is equal to four times three negative four times three so that square root of negative four can be two I square root of three then divided by 2.
So that here at the end we have y2 which equals to this will be -1 + i square root of 3.
This is y2.
Then here y3 will be -1 i square root of 3.
And these are three solutions that we get from this question. So here we have to check if the answer is correct because we have given y cubed over 2 equal to 4. And we have seen one real solution which is equals to 2 right here.
So that this is what we are going to replace to get 2 power 3 over 2 will be 4 {question mark}.
This will be 8 over 2 which is equal to 4. So that 4 equal to 4.
Left uh left-hand side equals to right-hand side. It means that our answer is correct.
Thanks for watching. Don't forget to share and subscribe. Bye-bye.
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