[Blazy]

I've put this under Sixth Form but it's just an interesting question I found.

Given:




What is ?

Any neat way of doing this? I Seeing as it looks quite cyclical (not sure if this is the word), I assume there must be a neat way of doing this.

I considered some mindless algebra bashing via but it looks messy, though does come out quite nicely. I've looked at it in some kind of matrix form but nothing sticks out.

[rubbermat]

Without going into it too deeply, have you thought of the difference of squares and factoring from there?
That may be one approach.

[Intriguing Alias]

(Original post by Blazy)
I've put this under Sixth Form but it's just an interesting question I found.

Given:




What is ?

Any neat way of doing this? I Seeing as it looks quite cyclical (not sure if this is the word), I assume there must be a neat way of doing this.

I considered some mindless algebra bashing via but it looks messy, though does come out quite nicely. I've looked at it in some kind of matrix form but nothing sticks out.

Well at first glance, I'd say you could just work out p^2, simplify as far as possible and then q and r come very easily by swapping a for b, b for c etc. rather than squaring each one separately then adding together.

That's the best thing I can see so far - not sure about anything easier. But in fact; you could just work out p^2.

which gives you:



Then where you get ab, write ab + ac + bc (because you'll get ac + bc from q and r)

Then do the same with ab and bc, then you can write a^2 + b^2 + c^2 where we have a^, b^2 OR c^2 above.

To get:

Spoiler:

Show

Then that's basically done. You'll find it simplifies very nicely (well it looks like it will to me).


EDIT: I did the simplifying from where I got and it comes out VERY nicely:

Spoiler:

Show

[nuodai]

(Original post by hassi94)
EDIT: I did the simplifying from where I got and it comes out VERY nicely:

Spoiler:

Show

Something doesn't seem right here. The only terms contributing to the term are , and so the coefficient of should be .

I reckon there's an easier way around this than expansion and simplification. For instance, we know that for some since is an even quartic (it's obviously quartic, and we can check that it's even by considering signs of terms without needing to calculate it explicitly). A bit more rummaging in this direction might lead you to an answer.

[Intriguing Alias]

(Original post by nuodai)
Something doesn't seem right here. The only terms contributing to the term are , and so the coefficient of should be .

I reckon there's an easier way around this than expansion and simplification. For instance, we know that for some since is an even quartic (it's obviously quartic, and we can check that it's even by considering signs of terms without needing to calculate it explicitly). A bit more rummaging in this direction might lead you to an answer.

Yep sorry - that should've been a^2 where I wrote a and so on


Despite the error though, that took me a fairly small amount of time, I started just after making the initial post and finished at about 20 to 11.

[nuodai]

(Original post by hassi94)
Yep sorry - that should've been a^2 where I wrote a and so on

Ah right, that seems right in that case.

I've worked out an alternative solution following from what I put in my last post.

Spoiler:

Show

We know that . And by considering the constant terms we must have





(This expansion was easy by symmetry in .)

And by substituting we get



which gives .

Hence

[Intriguing Alias]

(Original post by nuodai)
Ah right, that seems right in that case.

I've worked out an alternative solution following from what I put in my last post.

Spoiler:

Show

We know that . And by considering the constant terms we must have





(This expansion was easy by symmetry in .)

And by substituting we get



which gives .

Hence

Nice one - a lot harder to make a mistake in yours, that's certain

[Blazy]

(Original post by nuodai)
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(Original post by hassi94)
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I see, so it's worth trying to see what terms you'd get out - the comparing the coefficient one is pretty damn neat. Thanks!

PRSOM.