Question 5(v)
Show that:
(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

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Solution

L.H.S = $(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a)$
$= a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}$ $[U s i n g i d e n t i t y (a - b) (a + b) = a^{2} - b^{2}]$
=0.
=R.H.S

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Question 5(v)
Show that:
(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

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

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

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

(a + b) (a - b) + (b + c) (b - c) + (c + a) (c - a)

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