Question

(a²-b²)³+(b²+c³)³+(c²-a²)³
(a-b)³+(b-c)³+(c-a)³
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Answer 1

Answer:

Step-by-step explanation:

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.

Answer 2

Step-by-step explanation:

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

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...

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 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) = 0we assume that   a ≠b ≠ c.

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) = 0we assume that   a ≠b ≠ c.hence,

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) = 0we assume that   a ≠b ≠ c.hence,(a²-b²)³+(b²-c²)³+(c²-a²)³  ÷  (a-b)³+b-c)³+(c-a)³

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) = 0we 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) ]

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) = 0we 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)

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) = 0we 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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