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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 $(a+b)(b+c)(a+c)=0$
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)

@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 .

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)

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.

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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 $c^{2n+1} = c^{2n+1}$

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

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See my approach:

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

Am not able to do more than this...

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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).

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

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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 .

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@Kalash Verma

Thanks for your help! :)Log in to reply

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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)

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$a,b,c \neq 0$.

Yeah , my method hasLog in to reply

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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.

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

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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.

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Induction on $n$ ??

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Most probably .. xD

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

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