Show that
(a-b) (a+b) +(b-c) (b+c) +(c-a) (c+a) =0
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(a-b)(a+b) = a2 - b2
similarly.....
a2 - b2 + b2 - c2 + c2 - a = 0
a2 - a2 + b2 - b2 + c2 - c2 = 0
0+ 0+0 = 0
similarly.....
a2 - b2 + b2 - c2 + c2 - a = 0
a2 - a2 + b2 - b2 + c2 - c2 = 0
0+ 0+0 = 0
Answered by
1
yes for sure it's the right answer you can prove by keeping A as 1 and B as 2 and C as 3. now u can see that there is
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