Question

a³(b-c)³+b³(c-a)³+c³(a-b)³=3abc(a-b)(b-c)(c-a)​

Answer 1

Step-by-step explanation:

A³(b - c)³ + b³(c - a)³ + c³(a - b)³

= {a(b - c)}³ + {b( c - a)}³ + {c(a - b)}³

we see that,

a(b - c) + b(c - a) + c ( a - b) = 0

so, we can use ,

x³ + y³ + z³ = 3xyz when, (x + y + z) = 0

hence,

{a(b - c)}³ + {b(c - a)}³ + {c(a - b)}³ = 3abc(b - c)(c -a )(a - b)

Answer 2

Answer:

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