B Super silly question about a polynomial identity

[greg_rack]

Am I(always) legitimized to write $-(a-b)^n=(b-a)^n$?
I don't know why but it's confusing me... can't really understand when and why I can use that identity

 

[fresh_42]

Simply try some low numbers for $n$. And of course, it is allow in case $a=b$.
Or use the rule $(\alpha\cdot \beta)^n=\alpha^n\cdot \beta^n.$

 

[greg_rack]

Simply try some low numbers for $n$. And of course, it is allow in case $a=b$.

Yup, it works for $n=2$ and $n=3$, so I'd say it could be generalized to any odd and even integer index...

I imagine the most rigorous demonstration for it would be done by using Newton's binomial theorem(?), but is there an intuitive/more-straightforward way to demonstrate it, in order to get me convinced? :)

 

[fresh_42]

How can it work for $n=2$ if squares are non negative, but the left hand side is? Use the distribution law and pull $(-1)^n$ out.

 

[phinds]

Yup, it works for $n=2$

So, for example it would be true that -(2-1)*2 = (1-2)*2, yes?

EDIT: I see fresh beat me to it

 

[greg_rack]

How can it work for $n=2$ if squares are non negative, but the left hand side is?

Nevermind, I wrote without thinking...

By the way:
$$(b-a)^n=[-1(a-b)]^n=(-1)^n(a-b)^n$$
which is equal to $-(a-b)^n$ only for odd indexes.

 

[fresh_42]

Nevermind, I wrote without thinking...

By the way:
$$(b-a)^n=[-1(a-b)]^n=(-1)^n(a-b)^n$$
which is equal to $-(a-b)^n$ only for odd indexes.

Yes, you have $-(a-b)^n=(-1)^n(a-b)^n$. Now comes the important step which is usually forgotten and a never ending well of mistakes!

Case 1: $(a-b)^n = 0 \Longrightarrow a=b$
Case 2: $(a-b)^n\neq 0 \Longrightarrow (-1)^n=-1 \Longrightarrow n \text{ odd }$

 

[greg_rack]

Yes, you have $-(a-b)^n=(-1)^n(a-b)^n$. Now comes the important step which is usually forgotten and a never ending well of mistakes!

Case 1: $(a-b)^n = 0 \Longrightarrow a=b$
Case 2: $(a-b)^n\neq 0 \Longrightarrow (-1)^n=-1 \Longrightarrow n \text{ odd }$

Got it, thanks a lot!

 

[Vanadium 50]

Yup, it works for n = 2 and n = 3

Are you sure? Are you sure you're sure?