Math, asked by mepankajrar, 1 year ago

(a2-b2) +(b2-c2)+(c2-a2)/(a-b)2+(b-c)2+(c-a)2

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Answered by pulkitkumar1010
2

1 is its right answer

Answered by Anonymous
5


1


(a²-b²)³+(b²-c²)³+(c²-a²)³÷(a-b)³+b-c)³+(c-a)³=

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Ask for details FollowReport byKhushimudiraj1 12.09.2016

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kvnmurty

we use the fact that if a+b+c = 0, then a³+b³+ c³ = 3 a b c.

(a² - b²) + (b² - c²) + (c² - a²)  = 0...
also   (a-b) + (b-c)+ (c-a) = 0
we assume that   a ≠b ≠ c.

hence,
(a²-b²)³+(b²-c²)³+(c²-a²)³  ÷  (a-b)³+b-c)³+(c-a)³
=  3 (a² - b²) (b² - c²) (c² - a²)  / [ 3 (a - b) (b - c) ( c -a) ]
=  ( a+b) (b + c) ( c +a)

∵  a² - b²  = (a -b) (a+b)   and similarly other terms.
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