show that, (a-b)(a+b)+(b-c)(b+c)+(c-a)(c-b)=0
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Hello friend!
Here's ur answer!
(a-b)(a+b)+(b-c)(b+c)+(c-a)(c-b)=0
= (a2 - b2 )+( b2 - C2) + (c2- a2)
= a 2- a 2 + b2- b2+ c2 - c2
= 0
Here's ur answer!
(a-b)(a+b)+(b-c)(b+c)+(c-a)(c-b)=0
= (a2 - b2 )+( b2 - C2) + (c2- a2)
= a 2- a 2 + b2- b2+ c2 - c2
= 0
Geniusman:
new logic very nice
Is there any other logic?
a little different but your one is acceptable
Ok
Can u tell me that method in msg box?
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