(a2-b2) +(b2-c2)+(c2-a2)/(a-b)2+(b-c)2+(c-a)2
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(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.
1
(a²-b²)³+(b²-c²)³+(c²-a²)³÷(a-b)³+b-c)³+(c-a)³=
Advertisement
Ask for details FollowReport byKhushimudiraj1 12.09.2016
Answers

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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