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How to solve this Q???

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Reply 1
Avatar for Andy98
Andy98
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Original post by Omarsa1


Right ok, have you managed to start it?

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Duplicate thread.

http://www.thestudentroom.co.uk/showthread.php?t=3794079
Reply 3
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Omarsa1
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I just made a new account so I posted it again


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Reply 4
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Omarsa1
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I managed to expand it but couldn't do the show that


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Reply 5
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Andy98
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Original post by Omarsa1
I managed to expand it but couldn't do the show that


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And what expansion did you end up with?

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Reply 6
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Omarsa1
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Ended with that ImageUploadedByStudent Room1451223261.324018.jpg


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Original post by Omarsa1
Ended with that ImageUploadedByStudent Room1451223261.324018.jpg


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That's not quite right. You should have x3+3x+3x+1x3x^3+3x+\frac{3}{x}+\frac{1}{x^3}
Reply 8
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Omarsa1
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You are right


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Reply 9
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Original post by BuryMathsTutor
That's not quite right. You should have x3+3x+3x+1x3x^3+3x+\frac{3}{x}+\frac{1}{x^3}


You are right but how to show that it's =18


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Reply 10
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Andy98
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Original post by Omarsa1
You are right but how to show that it's =18


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Well how would you solve a quadratic? Equate it to 0, right? Using that logic, equate this one to 0. Then rearrange so that the left hand side matches the left hand side of the one in the show that. You know the right hand side must equal 18 so solve whatever ends up on the right hand side of the equation you just rearranged equal to 18. Substitute the value you get for x into the equation from the question

Original post by BuryMathsTutor
That's not quite right. You should have x3+3x+3x+1x3x^3+3x+\frac{3}{x}+\frac{1}{x^3}


Quoting you so I can see the equation

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Original post by Omarsa1
You are right but how to show that it's =18


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Write it as x3+1x3+3(x+1x)x^3+\frac{1}{x^3}+3(x+\frac{1}{x}).
Reply 12
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davros
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Original post by Andy98
Well how would you solve a quadratic? Equate it to 0, right? Using that logic, equate this one to 0. Then rearrange so that the left hand side matches the left hand side of the one in the show that. You know the right hand side must equal 18 so solve whatever ends up on the right hand side of the equation you just rearranged equal to 18. Substitute the value you get for x into the equation from the question

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This isn't a quadratic, it's a cubic!

And you're not trying to solve for x, you're supposed to be using the information you're given, plus a rearrangement of the binomial expansion, to show that the desired expression is equal to 18. Solving for x is NOT the way to go here :smile:
Reply 13
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Omarsa1
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I can't solve it


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Reply 14
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Original post by davros
This isn't a quadratic, it's a cubic!

And you're not trying to solve for x, you're supposed to be using the information you're given, plus a rearrangement of the binomial expansion, to show that the desired expression is equal to 18. Solving for x is NOT the way to go here :smile:


Can't get 18 can you get


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Reply 15
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Andy98
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Original post by davros
This isn't a quadratic, it's a cubic!

And you're not trying to solve for x, you're supposed to be using the information you're given, plus a rearrangement of the binomial expansion, to show that the desired expression is equal to 18. Solving for x is NOT the way to go here :smile:


I'm just saying how I'd do it - I know it's not a quadratic, although the equation in the show that part is a hidden quadratic. But I would rearrange the binomial expansion, then you'd have to solve 18=−3x−3x18= -3x - \frac{3}{x} in order to prove the other bit. Or have I been doing maths wrong my entire life?
Reply 16
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Zacken
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Original post by Andy98
I'm just saying how I'd do it - I know it's not a quadratic, although the equation in the show that part is a hidden quadratic. But I would rearrange the binomial expansion, then you'd have to solve 18=−3x−3x18= -3x - \frac{3}{x} in order to prove the other bit. Or have I been doing maths wrong my entire life?


You don't want to find xx.
Reply 17
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Andy98
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Original post by Zacken
You don't want to find xx.


But you have to show that the equation works, and I can't imagine it being proof by induction
Original post by BuryMathsTutor
That's not quite right. You should have x3+3x+3x+1x3x^3+3x+\frac{3}{x}+\frac{1}{x^3}


Yeah so that s&it is now equal to 27, since we know it's equal to (x+1/x)^3 and x+1/x is 3 so the whole thing is 3^3=27.
Then take 3 out and solve for what you need, as other people have said.
Reply 19
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Zacken
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Original post by Andy98
But you have to show that the equation works, and I can't imagine it being proof by induction

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