\[\frac {x-b^2}{a+c} + \frac {x-c^2}{a+b} + \frac {x-a^2}{b+c} = 4(a+b+c)\]

How can I solve this equation for \(x\)? Please show me the process and also the basic idea of solving this type of equation.

\[\frac {x-b^2}{a+c} + \frac {x-c^2}{a+b} + \frac {x-a^2}{b+c} = 4(a+b+c)\]

How can I solve this equation for \(x\)? Please show me the process and also the basic idea of solving this type of equation.

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TopNewestx=(a+b+c)^2 – Md Hasib · 8 months, 3 weeks ago

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What have you tried? Where did you get stuck?

Given that this is a linear equation in \(x\), it should be simple to shift terms around. (Of course, there may be other approaches.)

There is no guarantee that the answer you obtain will be "nice". Though, if you saw it in an olympiad, then it would likely be nice. – Calvin Lin Staff · 8 months, 3 weeks ago

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– Omar Sayeed Saimum · 8 months, 3 weeks ago

My answer is not "nice". A big one. But it's an Olympiad question according to my friend ( He gave me this problem to solve). So.....Log in to reply

– Calvin Lin Staff · 8 months, 3 weeks ago

Can you state your answer? Ideally, make it as simplified as possible.Log in to reply

– Omar Sayeed Saimum · 8 months, 3 weeks ago

X= (4(a+b+c)^2(ab+bc+ca) + (a+b+c)(a^3+b^3+c^3))/((a+b+c)^2 + (ab+bc+ca))Log in to reply

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– Omar Sayeed Saimum · 8 months, 3 weeks ago

Please look at that again. the total term is divided by ((a+b+c)^2 + (ab+bc+ca)).Log in to reply

I agree with the denominator being \( (a+b)(b+c) + (b+c)(c+a) + (c+a)(a+b) \). It is more natural to write it as \( a^2 + b^2 + c^3 + 3ab + 3bc + 3ca \).

I'm not sure how you arrived at the numerator though. To me, it should be \( b^2(b+c)(b+a) + c^2 (c+a)(c+b) + a^2(a+b)(a+c) + 4(a+b+c)(a+b)(b+c)(c+a) \).

As always, if you lay out your steps, someone would be able to see what you did and provide feedback from there. But otherwise, we can only guess at what you're doing. – Calvin Lin Staff · 8 months, 3 weeks ago

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– Omar Sayeed Saimum · 8 months, 3 weeks ago

Yeah, the denominator is a^2 + b^2 + c^2 + 3(ab+bc+ca) I've just separated the term using (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca) Anyway, if you prefer to lay out all my steps, then I'll do so. Oh, I think the answer is (a+b+c)^2. I was using some terms in the equation randomly. Then suddenly I noticed that if we use x = (a+b+c)^2, the equation becomes true.Log in to reply

Other than making a lucky observation about the answer, how else can we appraoch this problem? – Calvin Lin Staff · 8 months, 3 weeks ago

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– Omar Sayeed Saimum · 8 months, 3 weeks ago

That's what I'm asking to you. Still I'm trying to get this. But can't find anyway. Wil you kindly show me the process if you get that? Then I'll be relieved from this headache.😊Log in to reply

Here are some observations:

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– Omar Sayeed Saimum · 8 months, 3 weeks ago

Thanks. But I didn't understand the 3rd line of the 5th step. x = (a+b+c)^2 (mod(a+b)(b+c)(c+a)) How?Log in to reply

It is clear that \( x = (a+b+c)^2 \) satisfies \( x = c^2 + k_c (a+b) = a^2 + k_a (b+c) = b^2 + k_b( c+a) \), but it's not clear how we can easily backtrack. – Calvin Lin Staff · 8 months, 3 weeks ago

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## Calvin Lin plz help me.

– Omar Sayeed Saimum · 8 months, 3 weeks agoLog in to reply

– Pi Han Goh · 8 months, 3 weeks ago

Instead of asking people for help, tell us where you're stuck because we don't know what issues you're facing.Log in to reply

– Omar Sayeed Saimum · 8 months, 3 weeks ago

I didn't solve this kind of equation before. I'm trying. But couldn't find any way. So please help me to solve this.Log in to reply