Factorise: a³(b-c)³+b³(c-a)³+c³(a-b)³
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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)
= {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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answer=3abc (b-c) (c-a) (a-b)
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