Chapter (Factorization)
The sum or difference of 2 cubes (Title):
Equation:
a ^4 + 4b^4 + 3a^2b^2
A step-by-step answer would be appreciated a lot!
Thanks
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Math Sum Help!!!!
Chapter (Factorization)
The sum or difference of 2 cubes (Title):
Equation:
a ^4 + 4b^4 + 3a^2b^2
A step-by-step answer would be appreciated a lot!
Thanks
Are you sure you've written down the question correctly?Originally Posted by ArdaDelight
As it stands it's not a sum/difference of two cubes.
It looks like it could be a quadratic trinomial of a fourth degree if you rearrange the 2nd and 3rd components. But as a quadratic trinomial it can't be factorised.
The best I can do for you is by grouping:
(a^4 +3a^2b^2) +4b^4 or a^4 +(3a^2b^2 +4b^4)
you get
a^2(a^2 +3b^2) +4b^4 or a^4 + b^2(3a^2 +4b^2)
Oh, by the way... it's not an equation, it's an expression.
Thanks for the effort mate, appreciate it!Are you sure you've written down the question correctly?
As it stands it's not a sum/difference of two cubes.
It looks like it could be a quadratic trinomial of a fourth degree if you rearrange the 2nd and 3rd components. But as a quadratic trinomial it can't be factorised.
The best I can do for you is by grouping:
(a^4 +3a^2b^2) +4b^4 or a^4 +(3a^2b^2 +4b^4)
you get
a^2(a^2 +3b^2) +4b^4 or a^4 + b^2(3a^2 +4b^2)
Oh, by the way... it's not an equation, it's an expression.
I have the answers at the back of my TB, but I need to show the steps, in case this would help;
The answer I'm supposed to get is: (a^2 + ab + 2b^2) (a^2 - ab + 2b^2)
No foreign language on the forums please.Are you sure you've written down the question correctly?
As it stands it's not a sum/difference of two cubes.
It looks like it could be a quadratic trinomial of a fourth degree if you rearrange the 2nd and 3rd components. But as a quadratic trinomial it can't be factorised.
The best I can do for you is by grouping:
(a^4 +3a^2b^2) +4b^4 or a^4 +(3a^2b^2 +4b^4)
you get
a^2(a^2 +3b^2) +4b^4 or a^4 + b^2(3a^2 +4b^2)
Oh, by the way... it's not an equation, it's an expression.
Lol, this is 9th grade Maths these days!No foreign language on the forums please.
Yes, the answer is consistent with the question, it's getting to the answer that's the problem.
To start with try breaking up the 3a^2b^2 into a^2b^2 + 2a^2b^2 and try factorising through grouping common factors.
I've started doing that, but I haven't reached the answer yet.
No foreign language on the forums please.
Last edited by tweepie; 17-9-11 at 16:24. Reason: spelling
"Twelfty"
I did maths to O'Level and managed a pass but this stuff is definitely foreign to me, was a long time ago mind you.Yes, the answer is consisitent with the question, it's getting to the answer that's the problem.
To start with try breaking up the 3a^2b^2 into a^2b^2 + 2a^2b^2 and try factorising through grouping common factors.
I've started doing that, but I haven't reached the answer yet.
Hi, i would go about it like this:
a^4 + 4b^4 + 3(a^2)(b^2)
= (a^2)^2 + (2b^2)^2 + 3(ab)^2
=(a^2 + 2b^2)^2 - (ab)^2
=(a^2 + ab + 2b^2)(a^2 - ab + 2b^2) since {x^2 - y^2
= (x - y)(x+y)
you could also let x = a^2 and y= b^2. Makes it easier to look at
Hope i helped. Cheers
Last edited by LMYS; 17-9-11 at 16:32.
Actually, (x + y)(x - y) = x^2 - y^2.Hi, i would go about it like this:
a^4 + 4b^4 + 3(a^2)(b^2)
= (a^2)^2 + (2b^2)^2 + 3(ab)^2
=(a^2 + 2b^2)^2 - (ab)^2
=(a^2 + ab + 2b^2)(a^2 - ab + 2b^2) since {x^2 - y^2
= (x - y) ^2
you could also let x = a^2 and y= b^2. Makes it easier to look at
Hope i helped. Cheers
The first bit of your solution makes sense but you changed the last part of the expression for some reason.
Dont listen to Tweepie, She's a science teacher.
Thanks all you guys! Espescially the 2 above cause you got the answer.Hi, i would go about it like this:
a^4 + 4b^4 + 3(a^2)(b^2)
= (a^2)^2 + (2b^2)^2 + 3(ab)^2
=(a^2 + 2b^2)^2 - (ab)^2
=(a^2 + ab + 2b^2)(a^2 - ab + 2b^2) since {x^2 - y^2
= (x - y) ^2
you could also let x = a^2 and y= b^2. Makes it easier to look at
Hope i helped. Cheers
I'm going to get 20/20 on my internals.
Nyway thanks a lot!
changed it. Its correct now thanks for spotting that banglared. I meant to type that out. I guess my brain and fingers arent working in sync at the moment.this could be a problem.lol
no prob man, anything to help
Got it,
I could see it was difference of squares from your answer but I was approaching the squares from the wrong point.
Focus on the result that a^2 + b^2 is (a+b)^2 - 2ab
However, you are working with a^4 and 4b^4 instead (in reality (a^2)^2 and (2b^2)^2)
So breaking up your expression you have:
(a^2)^2 + (2b^2)^2 + 3a^2b^2
Then using the result above gives you:
(a^2 + 2b^2)^2 -2(a^2)(2b^2) +3a^2b^2
gives:
(a^2 + 2b^2)^2 - 4a^2b^2 +3a^2b^2
gives:
(a^2 +2b^2)^2 - a^2b^2
gives:
(a^2 +2b^2)^2 - (ab)^2
Now you have a difference of squares so apply your usual difference of squares method to get the final result!
It was you mentioning cubes put me off earlier.... I couldn't see where the cubes were coming from.
P.S. I've colour coded the parts to help you keep track.
Looks like someone got there before me!!!
Tweepie, I'm having trouble solving Fermat's last theorem. Any help would be appreciated.
I think that's where you're going wrong.
You can't solve a theorem, you should be trying to prove the theorem!!!
I think that theorem will never be proven.
Thanks mate.
Got it,
I could see it was difference of squares from your answer but I was approaching the squares from the wrong point.
Focus on the result that a^2 + b^2 is (a+b)^2 - 2ab
However, you are working with a^4 and 4b^4 instead (in reality (a^2)^2 and (2b^2)^2)
So breaking up your expression you have:
(a^2)^2 + (2b^2)^2 + 3a^2b^2
Then using the result above gives you:
(a^2 + 2b^2)^2 -2(a^2)(2b^2) +3a^2b^2
gives:
(a^2 + 2b^2)^2 - 4a^2b^2 +3a^2b^2
gives:
(a^2 +2b^2)^2 - a^2b^2
gives:
(a^2 +2b^2)^2 - (ab)^2
Now you have a difference of squares so apply your usual difference of squares method to get the final result!
It was you mentioning cubes put me off earlier.... I couldn't see where the cubes were coming from.
P.S. I've colour coded the parts to help you keep track.
Really appreciate the help that all you guys provided.
Are you sure you've written down the question correctly?
As it stands it's not a sum/difference of two cubes.
It looks like it could be a quadratic trinomial of a fourth degree if you rearrange the 2nd and 3rd components. But as a quadratic trinomial it can't be factorised.
The best I can do for you is by grouping:
(a^4 +3a^2b^2) +4b^4 or a^4 +(3a^2b^2 +4b^4)
you get
a^2(a^2 +3b^2) +4b^4 or a^4 + b^2(3a^2 +4b^2)
Oh, by the way... it's not an equation, it's an expression.
I thought it was forbidden to type in foreign languages on here?
It has been proven. By Andrew wiles.I think that's where you're going wrong.
You can't solve a theorem, you should be trying to prove the theorem!!!
I think that theorem will never be proven.
Anyone who's maths and/or programming geek should check out this site:
http://projecteuler.net/
I've been trying to do some when bored in work. Managed about 5 so far.
How does this equate to life in the real world!!I thought it was forbidden to type in foreign languages on here?
Where's Sam Whitwicky when you need him?
a^4 + 4b^4 + 3a^2b^2
= a^4 + 4b^4 + 4a^2b^2 - a^2b^2
= (a^2 + 2b^2)^2 - (ab)^2
Ahh im so glad i swapped A level maths for Psychology.
The answer's 6.
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