If \(\quad a^3+b^3+c^3 = (a+b+c)^3\) ,

Prove that \(\quad \large a^{2n+1}+ b^{2n+1}+ c^{2n+1} = (a+b+c)^{2n+1}\) for all \(n \in

N\)

Nice proofs are always welcome. :)

If \(\quad a^3+b^3+c^3 = (a+b+c)^3\) ,

Prove that \(\quad \large a^{2n+1}+ b^{2n+1}+ c^{2n+1} = (a+b+c)^{2n+1}\) for all \(n \in

N\)

Nice proofs are always welcome. :)

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TopNewestWell I THINK it can be proved without induction.@Nihar Mahajan @Vaibhav Prasad I will write it in short. It can easily be derived that \((a+b)(b+c)(a+c)=0\) Now we have 3 cases in first all the 3 terms are 0 , or any 2terms are 0 or only one term is 0.

Case\( (a+b)=0\)

\(a=-b\)

So now just substitute a = -b in the equation (which is to be proved)

That will give \(c^{2n+1} = c^{2n+1}\) – Kalash Verma · 1 year, 9 months ago

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– Nihar Mahajan · 1 year, 9 months ago

Oh ,,, I got your method... Thanks!Log in to reply

\[a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc \\ \Rightarrow 3(a+b+c)(ab+bc+ac) = 3abc \\ (a+b+c)\left(\dfrac{ab+bc+ac}{abc}\right)=1 \quad or \quad (a+b+c)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\]

Am not able to do more than this... – Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan

Take \(a+c=x\) Then \((b+x)(bx+ac) = abc\)

Multiplying and simplifying gives us \(x(b^{2}+bx+ac )= 0\)substitute x Then Further factorizing gives us \(a+b)(b+c)(a+c) = 0\)

Now the method goes as usual (given in my method). – Kalash Verma · 1 year, 9 months ago

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@Kalash Verma I understood your method. I am asking , can you extend my approach furthur? – Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan Well actually its the extension of ur approach. In the third line I have replaced a+c by x and then I have explained above.In the end both results in same factors . – Kalash Verma · 1 year, 9 months ago

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@Kalash Verma – Nihar Mahajan · 1 year, 9 months ago

Thanks for your help! :)Log in to reply

– Kalash Verma · 1 year, 9 months ago

Don't mention it.AND you have 703 followers and 703 following. Wow.congratsLog in to reply

– Kalash Verma · 1 year, 9 months ago

I will try using this method and tell you about it . But this method does not work when a or b or c=0 and if its right then it proves 0/0=1 (lol was it so easy)Log in to reply

– Nihar Mahajan · 1 year, 9 months ago

Yeah , my method has \(a,b,c \neq 0\).Log in to reply

– Kalash Verma · 1 year, 9 months ago

I will try solving it by your method and tell you tomorrow (well If I can solve it).Log in to reply

Well, if the first condition is met, then one of these are true, \(a+b=0\), \(b+c=0\), \(c+a=0\), and so the next condition follows immediately. – Michael Mendrin · 1 year, 9 months ago

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@Michael Mendrin @Calvin Lin Sir , can you extend my approach that I have written above? – Nihar Mahajan · 1 year, 9 months ago

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– Michael Mendrin · 1 year, 9 months ago

I can't go anywhere with that approach either. But if I just solve it for \(=0\), I get the condition \(a+b=0\), \(b+c=0\), or \(c+a=0\), and so the rest becomes easy.Log in to reply

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– Nihar Mahajan · 1 year, 9 months ago

Where has \(3abc\) gone?Log in to reply

Induction on \(n\) ?? – Vaibhav Prasad · 1 year, 9 months ago

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– Nihar Mahajan · 1 year, 9 months ago

Most probably .. xDLog in to reply

– Vaibhav Prasad · 1 year, 9 months ago

I always prefer induction... even though it's kind of a shortcutLog in to reply