Question

using suitable identity show that (a-b) (a+b) + (b-c) (b+c) + (c-a) (c+a) =0​

Answer 1

(a^2-b^2)+(b^2-c^2)+(c^2-a^2)

(a^2-a^2)+(b^2-b^2)+(c^2-c^2)

0+0+0=0

Answer 2

Answer:

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To prove :-

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

@nswer :-

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

$({a}^{2} - {b}^{2} ) + ({b }^{2} - {c}^{2}) + ( {c}^{2} - {a}^{2} )$

Using Identity :

$(a - b) \: (a + b) = {a}^{2} - {b}^{2}$

Now ,

Simplify ,

${a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2}$

$= 0$

Therefore , It is proved that LHS = RHS