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