Question

(a²-b²)³ + (b²-c²)³ +(c²-a²)³ /(a−b)³+(b-c)³+(c-a)³​

Answer 1

Step-by-step explanation:

we use the fact that if a + b + c = 0 then a ^ 3 + b ^ 3 + c ^ 3 = 3abc.

(a^ 2 -b^ 2 )+(b^ 2 -c^ 2 )+(c^ 2 -a^ 2 )=0... also (a - b) + (b - c) + (c - a) = 0 we assume that a #b # c.

hence,

(a^ 2 -b^ 2 )^ 3 +(b^ 2 -c^ 2 )^ 3 +(c^ 2 -a^ 2 )^ 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

Answer:

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.