Q6 of 157 Page 22

Show that

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

Solving L.H.S. first,


(a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a)


We know that


∵ (a2 – b2) = (a – b) × (a + b)


⇒ (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = (a2 – b2) + (b2 – c2) + (c2 – a2)


⇒ (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = 0


∵ L.H.S. = R.H.S.


Hence, proved.


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