we know that (a-b)^2= (b-a)^2

square root on both sides

= a-b=b-a

= (a-b)= -(a-b)

if (a-b)=x

then x= -x

but how can a number be equal to its additive inverse

same case with (a+b-c)^2=(-a-b+c)^2 or (a-b-c)^2=(-a+b+c)^2 etc

can anyone explain me this problem

Answer :

If we have  (  a - b )²  =  (  b  -  a )²

So after taking square root on both hand side , we get

$\pm$( a - b )  =  $\pm$ ( b -  a )

And

if we take negative sign common , we get

$\pm$( a - b )  =  $\mp$ ( a -  b )

So if we get

x  = -x

It is because of our starting assumption (  a - b )²  =  (  b  -  a )² 

And , If here we put a  = 3 and b  = 2 , then we get

( 3  - 2 )² = ( 2  - 3 )²

1² = ( - 1 )²

1 = 1

So as we have whole square on both hand side , we always get a positive value on both hand side .