Oh wow!

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. :)

#Algebra

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Comments

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 - 6 years, 1 month ago

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

Nihar Mahajan - 6 years, 1 month 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.

Michael Mendrin - 6 years, 1 month ago

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)

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

A Former Brilliant Member - 6 years, 1 month ago

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...

Nihar Mahajan - 6 years, 1 month 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)

A Former Brilliant Member - 6 years, 1 month ago

@A Former Brilliant Member – Yeah , my method has $a,b,c \neq 0$ .

Nihar Mahajan - 6 years, 1 month ago

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

A Former Brilliant Member - 6 years, 1 month ago

@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).

A Former Brilliant Member - 6 years, 1 month ago

@A Former Brilliant Member – @Kalash Verma I understood your method. I am asking , can you extend my approach furthur?

Nihar Mahajan - 6 years, 1 month ago

@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 .

A Former Brilliant Member - 6 years, 1 month ago

@A Former Brilliant Member – Thanks for your help! :) @Kalash Verma

Nihar Mahajan - 6 years, 1 month ago

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

A Former Brilliant Member - 6 years, 1 month ago

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

Nihar Mahajan - 6 years, 1 month ago

Induction on $n$ ??

Vaibhav Prasad - 6 years, 1 month ago

Most probably .. xD

Nihar Mahajan - 6 years, 1 month ago

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

Vaibhav Prasad - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$