Oh wow!

If a3+b3+c3=(a+b+c)3\quad a^3+b^3+c^3 = (a+b+c)^3 ,

Prove that a2n+1+b2n+1+c2n+1=(a+b+c)2n+1\quad \large a^{2n+1}+ b^{2n+1}+ c^{2n+1} = (a+b+c)^{2n+1} for all nNn \in N

Nice proofs are always welcome. :)

Note by Nihar Mahajan
4 years, 2 months ago

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Well, if the first condition is met, then one of these are true, a+b=0a+b=0, b+c=0b+c=0, c+a=0c+a=0, and so the next condition follows immediately.

Michael Mendrin - 4 years, 2 months ago

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@Michael Mendrin @Calvin Lin Sir , can you extend my approach that I have written above?

Nihar Mahajan - 4 years, 2 months ago

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

Michael Mendrin - 4 years, 2 months ago

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Well 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(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)=0

a=ba=-b

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

That will give c2n+1=c2n+1c^{2n+1} = c^{2n+1}

Kalash Verma - 4 years, 2 months ago

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See my approach:

a3+b3+c3=(a+b+c)33(a+b+c)(ab+bc+ac)+3abc3(a+b+c)(ab+bc+ac)=3abc(a+b+c)(ab+bc+acabc)=1or(a+b+c)(1a+1b+1c)=1a^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 - 4 years, 2 months ago

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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)

Kalash Verma - 4 years, 2 months ago

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@Kalash Verma Yeah , my method has a,b,c0a,b,c \neq 0.

Nihar Mahajan - 4 years, 2 months ago

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@Nihar Mahajan I will try solving it by your method and tell you tomorrow (well If I can solve it).

Kalash Verma - 4 years, 2 months ago

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@Nihar Mahajan

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

Multiplying and simplifying gives us x(b2+bx+ac)=0x(b^{2}+bx+ac )= 0substitute x Then Further factorizing gives us a+b)(b+c)(a+c)=0a+b)(b+c)(a+c) = 0

Now the method goes as usual (given in my method).

Kalash Verma - 4 years, 2 months ago

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@Kalash Verma @Kalash Verma I understood your method. I am asking , can you extend my approach furthur?

Nihar Mahajan - 4 years, 2 months ago

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@Nihar Mahajan @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 - 4 years, 2 months ago

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@Kalash Verma Thanks for your help! :) @Kalash Verma

Nihar Mahajan - 4 years, 2 months ago

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@Nihar Mahajan Don't mention it.AND you have 703 followers and 703 following. Wow.congrats

Kalash Verma - 4 years, 2 months ago

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Oh ,,, I got your method... Thanks!

Nihar Mahajan - 4 years, 2 months ago

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Induction on nn ??

Vaibhav Prasad - 4 years, 2 months ago

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Most probably .. xD

Nihar Mahajan - 4 years, 2 months ago

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I always prefer induction... even though it's kind of a shortcut

Vaibhav Prasad - 4 years, 2 months ago

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