Question

Show that
(a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0
Please kindly solve it on a piece of paper

Answer 1

hope this helps you.....

Answer 2

$Hey\:There$‼

I don't have a pen & paper with me right now. I'm texting here.
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(a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0

As we know,
(x + y) (x – y) = x² – y²

So, we have to apply the above identity in the given equation.

(a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = L.H.S.

= (a² – b²) + (b² – c²) + (c² – a² )

= a² – b² + b² – c² + c² – a²

= a² – a² + b² – b² + c² – c²

= 0 + 0 + 0

= 0

= R.H.S.
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